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Unformatted text preview: 5E-12(pp 736-745) 1/18/06 10:08 AM Page 736 CHAPTER 12 Bessel functions, which are used to model the vibrations of drumheads and cymbals, are defined as sums of infinite series in Section 12.8. Notice how closely the computergenerated models (which involve Bessel functions and cosine functions) match the photographs of a vibrating rubber membrane. Infinite Sequences and Series 5E-12(pp 736-745) 1/18/06 10:08 AM Page 737 Infinite sequences and series were introduced briefly in A Preview of Calculus in connection with Zeno’s paradoxes and the decimal representation of numbers. Their importance in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 12.10 in order 2 to integrate such functions as eϪx . (Recall that we have previously been unable to do this.) Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be familiar with the basic concepts of convergence of infinite sequences and series. Physicists also use series in another way, as we will see in Section 11.12. In studying fields as diverse as optics, special relativity, and electromagnetism, they analyze phenomena by replacing a function with the first few terms in the series that represents it. |||| 12.1 Sequences A sequence can be thought of as a list of numbers written in a definite order: a 1, a 2, a 3, a 4, . . . , a n, . . . The number a 1 is called the first term, a 2 is the second term, and in general a n is the nth term. We will deal exclusively with infinite sequences and so each term a n will have a successor a nϩ1 . Notice that for every positive integer n there is a corresponding number a n and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write a n instead of the function notation f ͑n͒ for the value of the function at the number n. NOTATION ■ The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by ͕a n ͖ ϱ ͕a n ͖ n෇1 or EXAMPLE 1 Some sequences can be defined by giving a formula for the nth term. In the following examples we give three descriptions of the sequence: one by using the preceding notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that n doesn’t have to start at 1. (a) (b) ͭ ͮ ͭ ͮ n nϩ1 ϱ an ෇ n nϩ1 an ෇ ͑Ϫ1͒n͑n ϩ 1͒ 3n n෇1 ͑Ϫ1͒n͑n ϩ 1͒ 3n ͭ ͭ ͮ 1 2 3 4 n , , , ,..., ,... 2 3 4 5 nϩ1 ͮ 2 3 4 5 ͑Ϫ1͒n͑n ϩ 1͒ Ϫ , ,Ϫ , ,..., ,... 3 9 27 81 3n 737 5E-12(pp 736-745) 738 ❙❙❙❙ 1/18/06 10:08 AM Page 738 CHAPTER 12 INFINITE SEQUENCES AND SERIES (c) (d) ϱ {sn Ϫ 3 }n෇3 ͭ ͮ n␲ cos 6 {0, 1, s2, s3, . . . , sn Ϫ 3, . . .} a n ෇ sn Ϫ 3, n ജ 3 ϱ a n ෇ cos n෇0 ͭ n␲ , nജ0 6 1, ͮ n␲ s3 1 , , 0, . . . , cos ,... 2 2 6 EXAMPLE 2 Find a formula for the general term a n of the sequence ͭ ͮ 3 4 5 6 7 ,Ϫ , ,Ϫ , ,... 5 25 125 625 3125 assuming that the pattern of the first few terms continues. SOLUTION We are given that a1 ෇ 3 5 a2 ෇ Ϫ 4 25 a3 ෇ 5 125 a4 ෇ Ϫ 6 625 a5 ෇ 7 3125 Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator n ϩ 2. The denominators are the powers of 5, so a n has denominator 5 n. The signs of the terms are alternately positive and negative, so we need to multiply by a power of Ϫ1. In Example 1(b) the factor ͑Ϫ1͒ n meant we started with a negative term. Here we want to start with a positive term and so we use ͑Ϫ1͒ nϪ1 or ͑Ϫ1͒ nϩ1. Therefore, a n ෇ ͑Ϫ1͒ nϪ1 nϩ2 5n EXAMPLE 3 Here are some sequences that don’t have a simple defining equation. (a) The sequence ͕pn ͖, where pn is the population of the world as of January 1 in the year n. (b) If we let a n be the digit in the nth decimal place of the number e, then ͕a n ͖ is a welldefined sequence whose first few terms are ͕7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .͖ (c) The Fibonacci sequence ͕ fn ͖ is defined recursively by the conditions f1 ෇ 1 f2 ෇ 1 fn ෇ fnϪ1 ϩ fnϪ2 nജ3 Each term is the sum of the two preceding terms. The first few terms are ͕1, 1, 2, 3, 5, 8, 13, 21, . . .͖ This sequence arose when the 13th-century Italian mathematician known as Fibonacci solved a problem concerning the breeding of rabbits (see Exercise 65). a¡ 0 FIGURE 1 1 2 a™ a£ a¢ 1 A sequence such as the one in Example 1(a), a n ෇ n͑͞n ϩ 1͒, can be pictured either by plotting its terms on a number line as in Figure 1 or by plotting its graph as in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates ͑1, a1 ͒ ͑2, a2 ͒ ͑3, a3 ͒ ... ͑n, a n ͒ ... 5E-12(pp 736-745) 1/18/06 10:08 AM Page 739 SECTION 12.1 SEQUENCES an 739 From Figure 1 or 2 it appears that the terms of the sequence a n ෇ n͑͞n ϩ 1͒ are approaching 1 as n becomes large. In fact, the difference 1 1Ϫ 7 a¶= 8 0 ❙❙❙❙ n 1 2 3 4 5 6 7 n 1 ෇ nϩ1 nϩ1 can be made as small as we like by taking n sufficiently large. We indicate this by writing FIGURE 2 lim nlϱ n ෇1 nϩ1 In general, the notation lim a n ෇ L nlϱ means that the terms of the sequence ͕a n ͖ approach L as n becomes large. Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 4.4. 1 Definition A sequence ͕a n ͖ has the limit L and we write lim a n ෇ L nlϱ or a n l L as n l ϱ if we can make the terms a n as close to L as we like by taking n sufficiently large. If lim n l ϱ a n exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have the limit L. an an L L FIGURE 3 Graphs of two sequences with lim an= L 0 0 n n n ` A more precise version of Definition 1 is as follows. 2 Definition A sequence ͕an ͖ has the limit L and we write lim an ෇ L |||| Compare this definition with Definition 4.4.5. or nlϱ a n l L as n l ϱ if for every ␧ Ͼ 0 there is a corresponding integer N such that Խa n Խ ϪL Ͻ␧ whenever n Ͼ N 5E-12(pp 736-745) 740 ❙❙❙❙ 1/18/06 10:08 AM Page 740 CHAPTER 12 INFINITE SEQUENCES AND SERIES Definition 2 is illustrated by Figure 4, in which the terms a 1 , a 2 , a 3 , . . . are plotted on a number line. No matter how small an interval ͑L Ϫ ␧, L ϩ ␧͒ is chosen, there exists an N such that all terms of the sequence from a Nϩ1 onward must lie in that interval. a¡ FIGURE 4 a£ a™ aˆ 0 aN+1 aN+2 L-∑ L a˜ aß a∞ a¢ a¶ L+∑ Another illustration of Definition 2 is given in Figure 5. The points on the graph of ͕an ͖ must lie between the horizontal lines y ෇ L ϩ ␧ and y ෇ L Ϫ ␧ if n Ͼ N . This picture must be valid no matter how small ␧ is chosen, but usually a smaller ␧ requires a larger N. y y=L+∑ L y=L-∑ FIGURE 5 0 1 2 3 4 N n Comparison of Definition 2 and Definition 4.4.5 shows that the only difference between lim n l ϱ a n ෇ L and lim x l ϱ f ͑x͒ ෇ L is that n is required to be an integer. Thus, we have the following theorem, which is illustrated by Figure 6. 3 Theorem If lim x l ϱ f ͑x͒ ෇ L and f ͑n͒ ෇ a n when n is an integer, then lim n l ϱ an ෇ L. y y=ƒ L FIGURE 6 0 x 1 2 3 4 In particular, since we know that limx l ϱ ͑1͞x r ͒ ෇ 0 when r Ͼ 0 (Theorem 4.4.4), we have 1 lim ෇0 if r Ͼ 0 4 nl ϱ nr If a n becomes large as n becomes large, we use the notation lim n l ϱ a n ෇ ϱ. The following precise definition is similar to Definition 4.4.7. 5 Definition lim n l ϱ an ෇ ϱ means that for every positive number M there is an integer N such that an Ͼ M whenever n Ͼ N If lim n l ϱ a n ෇ ϱ, then the sequence ͕a n ͖ is divergent but in a special way. We say that ͕a n ͖ diverges to ϱ. 5E-12(pp 736-745) 1/18/06 10:08 AM Page 741 SECTION 12.1 SEQUENCES ❙❙❙❙ 741 The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar. If ͕a n ͖ and ͕bn ͖ are convergent sequences and c is a constant, then Limit Laws for Sequences lim ͑a n ϩ bn ͒ ෇ lim a n ϩ lim bn nlϱ nlϱ nlϱ lim ͑a n Ϫ bn ͒ ෇ lim a n Ϫ lim bn nlϱ nlϱ nlϱ lim ca n ෇ c lim a n nlϱ lim c ෇ c nlϱ nlϱ lim ͑a n bn ͒ ෇ lim a n ؒ lim bn nlϱ nlϱ lim lim a n an nlϱ ෇ bn lim bn nlϱ nlϱ if lim bn nlϱ nlϱ [ lim a np ෇ lim a n nlϱ nlϱ p 0 if p Ͼ 0 and a n Ͼ 0 The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7). Squeeze Theorem for Sequences If a n ഛ bn ഛ cn for n ജ n 0 and lim a n ෇ lim cn ෇ L, then lim bn ෇ L. nlϱ nlϱ nlϱ cn Another useful fact about limits of sequences is given by the following theorem, whose proof is left as Exercise 69. bn an 0 6 Խ Խ If lim a n ෇ 0, then lim a n ෇ 0. Theorem nlϱ nlϱ n FIGURE 7 The sequence ͕ bn ͖ is squeezed between the sequences ͕ a n ͖ and ͕ cn ͖. EXAMPLE 4 Find lim nlϱ n . nϩ1 SOLUTION The method is similar to the one we used in Section 4.4: Divide numerator and denominator by the highest power of n and then use the Limit Laws. lim nlϱ |||| This shows that the guess we made earlier from Figures 1 and 2 was correct. n ෇ lim nlϱ nϩ1 ෇ 1 1ϩ 1 n 1 ෇1 1ϩ0 Here we used Equation 4 with r ෇ 1. lim 1 nlϱ ෇ lim 1 ϩ lim nlϱ nlϱ 1 n 5E-12(pp 736-745) 742 ❙❙❙❙ 1/18/06 10:08 AM Page 742 CHAPTER 12 INFINITE SEQUENCES AND SERIES ln n . n EXAMPLE 5 Calculate lim nlϱ SOLUTION Notice that both numerator and denominator approach infinity as n l ϱ. We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to functions of a real variable. However, we can apply l’Hospital’s Rule to the related function f ͑x͒ ෇ ͑ln x͒͞x and obtain lim xlϱ ln x 1͞x ෇ lim ෇0 xlϱ 1 x Therefore, by Theorem 3 we have lim nlϱ an ln n ෇0 n EXAMPLE 6 Determine whether the sequence a n ෇ ͑Ϫ1͒n is convergent or divergent. 1 SOLUTION If we write out the terms of the sequence, we obtain 0 1 2 3 4 ͕Ϫ1, 1, Ϫ1, 1, Ϫ1, 1, Ϫ1, . . .͖ n _1 The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and Ϫ1 infinitely often, a n does not approach any number. Thus, lim n l ϱ ͑Ϫ1͒n does not exist; that is, the sequence ͕͑Ϫ1͒n ͖ is divergent. FIGURE 8 |||| The graph of the sequence in Example 7 is shown in Figure 9 and supports the answer. EXAMPLE 7 Evaluate lim nlϱ ͑Ϫ1͒n if it exists. n SOLUTION an 1 lim nlϱ Ϳ Ϳ ͑Ϫ1͒n n ෇ lim nlϱ 1 ෇0 n Therefore, by Theorem 6, 0 1 n lim nlϱ _1 FIGURE 9 ͑Ϫ1͒n ෇0 n EXAMPLE 8 Discuss the convergence of the sequence a n ෇ n!͞n n, where n! ෇ 1 ؒ 2 ؒ 3 ؒ и и и ؒ n. SOLUTION Both numerator and denominator approach infinity as n l ϱ but here we have no corresponding function for use with l’Hospital’s Rule (x! is not defined when x is not an integer). Let’s write out a few terms to get a feeling for what happens to a n as n gets large: a1 ෇ 1 7 a2 ෇ an ෇ 1ؒ2 2ؒ2 a3 ෇ 1ؒ2ؒ3 3ؒ3ؒ3 1 ؒ 2 ؒ 3 ؒ иии ؒ n n ؒ n ؒ n ؒ иии ؒ n It appears from these expressions and the graph in Figure 10 that the terms are decreas- 5E-12(pp 736-745) 1/18/06 10:08 AM Page 743 ❙❙❙❙ SECTION 12.1 SEQUENCES |||| CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 8 can be graphed by entering the parametric equations x෇t ing and perhaps approach 0. To confirm this, observe from Equation 7 that ͩ 1 n an ෇ 2 ؒ 3 ؒ иии ؒ n n ؒ n ؒ иии ؒ n ͪ Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So 1 0 Ͻ an ഛ n y ෇ t!͞t t and graphing in dot mode starting with t ෇ 1, setting the t-step equal to 1. The result is shown in Figure 10. We know that 1͞n l 0 as n l ϱ. Therefore, a n l 0 as n l ϱ by the Squeeze Theorem. 1 EXAMPLE 9 For what values of r is the sequence ͕r n ͖ convergent? SOLUTION We know from Section 4.4 and the graphs of the exponential functions in Section 7.2 (or Section 7.4*) that lim x l ϱ a x ෇ ϱ for a Ͼ 1 and lim x l ϱ a x ෇ 0 for 0 Ͻ a Ͻ 1. Therefore, putting a ෇ r and using Theorem 2, we have 0 10 lim r n ෇ nlϱ FIGURE 10 ͭ ϱ 0 if r Ͼ 1 if 0 Ͻ r Ͻ 1 It is obvious that lim 1n ෇ 1 lim 0 n ෇ 0 and nlϱ nlϱ Խ Խ If Ϫ1 Ͻ r Ͻ 0, then 0 Ͻ r Ͻ 1, so Խ Խ Խ Խ lim r n ෇ lim r nlϱ nlϱ n ෇0 and therefore lim n l ϱ r n ෇ 0 by Theorem 6. If r ഛ Ϫ1, then ͕r n ͖ diverges as in Example 6. Figure 11 shows the graphs for various values of r. (The case r ෇ Ϫ1 is shown in Figure 8.) an an r>1 1 1 _1<r<0 0 r=1 n 1 0 FIGURE 11 The sequence an=r 743 1 0<r<1 n r<_1 n The results of Example 9 are summarized for future use as follows. The sequence ͕r n ͖ is convergent if Ϫ1 Ͻ r ഛ 1 and divergent for all other values of r. 8 lim r n ෇ nlϱ ͭ 0 1 if Ϫ1 Ͻ r Ͻ 1 if r ෇ 1 5E-12(pp 736-745) 744 ❙❙❙❙ 1/18/06 10:08 AM Page 744 CHAPTER 12 INFINITE SEQUENCES AND SERIES 9 Definition A sequence ͕an ͖ is called increasing if a n Ͻ a nϩ1 for all n ജ 1, that is, a1 Ͻ a2 Ͻ a3 Ͻ и и и . It is called decreasing if a n Ͼ a nϩ1 for all n ജ 1. It is called monotonic if it is either increasing or decreasing. EXAMPLE 10 The sequence ͭ ͮ 3 nϩ5 is decreasing because 3 3 3 Ͼ ෇ nϩ5 ͑n ϩ 1͒ ϩ 5 nϩ6 for all n ജ 1. (The right side is smaller because it has a larger denominator.) EXAMPLE 11 Show that the sequence a n ෇ n is decreasing. n2 ϩ 1 SOLUTION 1 We must show that a nϩ1 Ͻ a n , that is, nϩ1 n Ͻ 2 ͑n ϩ 1͒2 ϩ 1 n ϩ1 This inequality is equivalent to the one we get by cross-multiplication: nϩ1 n Ͻ 2 2 ͑n ϩ 1͒ ϩ 1 n ϩ1 &? ͑n ϩ 1͒͑n 2 ϩ 1͒ Ͻ n͓͑n ϩ 1͒2 ϩ 1͔ &? n 3 ϩ n 2 ϩ n ϩ 1 Ͻ n 3 ϩ 2n 2 ϩ 2n &? 1 Ͻ n2 ϩ n Since n ജ 1, we know that the inequality n 2 ϩ n Ͼ 1 is true. Therefore, a nϩ1 Ͻ a n and so ͕a n ͖ is decreasing. SOLUTION 2 Consider the function f ͑x͒ ෇ f Ј͑x͒ ෇ x : x ϩ1 2 x 2 ϩ 1 Ϫ 2x 2 1 Ϫ x2 ෇ 2 Ͻ0 2 2 ͑x ϩ 1͒ ͑x ϩ 1͒ 2 whenever x 2 Ͼ 1 Thus, f is decreasing on ͑1, ϱ͒ and so f ͑n͒ Ͼ f ͑n ϩ 1͒. Therefore, ͕a n ͖ is decreasing. 10 Definition A sequence ͕a n ͖ is bounded above if there is a number M such that an ഛ M for all n ജ 1 It is bounded below if there is a number m such that m ഛ an for all n ജ 1 If it is bounded above and below, then ͕a n ͖ is a bounded sequence. 5E-12(pp 736-745) 1/18/06 10:08 AM Page 745 SECTION 12.1 SEQUENCES an M L 0 1 23 FIGURE 12 n ❙❙❙❙ 745 For instance, the sequence a n ෇ n is bounded below ͑a n Ͼ 0͒ but not above. The sequence a n ෇ n͑͞n ϩ 1͒ is bounded because 0 Ͻ a n Ͻ 1 for all n. We know that not every bounded sequence is convergent [for instance, the sequence a n ෇ ͑Ϫ1͒n satisfies Ϫ1 ഛ a n ഛ 1 but is divergent from Example 6] and not every monotonic sequence is convergent ͑a n ෇ n l ϱ͒. But if a sequence is both bounded and monotonic, then it must be convergent. This fact is proved as Theorem 11, but intuitively you can understand why it is true by looking at Figure 12. If ͕a n ͖ is increasing and a n ഛ M for all n, then the terms are forced to crowd together and approach some number L. The proof of Theorem 11 is based on the Completeness Axiom for the set ‫ ޒ‬of real numbers, which says that if S is a nonempty set of real numbers that has an upper bound M (x ഛ M for all x in S ), then S has a least upper bound b. (This means that b is an upper bound for S, but if M is any other upper bound, then b ഛ M .) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line. 11 Monotonic Sequence Theorem Every bounded, monotonic sequence is convergent. Proof Suppose ͕a n ͖ is an increasing sequence. Since ͕a n ͖ is bounded, the set Խ S ෇ ͕a n n ജ 1͖ has an upper bound. By the Completeness Axiom it has a least upper bound L . Given ␧ Ͼ 0, L Ϫ ␧ is not an upper bound for S (since L is the least upper bound). Therefore aN Ͼ L Ϫ ␧ for some integer N But the sequence is increasing so a n ജ a N for every n Ͼ N . Thus, if n Ͼ N we have an Ͼ L Ϫ ␧ 0 ഛ L Ϫ an Ͻ ␧ so since a n ഛ L . Thus ԽL Ϫ a Խ Ͻ ␧ n whenever n Ͼ N so lim n l ϱ a n ෇ L . A similar proof (using the greatest lower bound) works if ͕a n ͖ is decreasing. The proof of Theorem 11 shows that a sequence that is increasing and bounded above is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.) This fact is used many times in dealing with infinite series. EXAMPLE 12 Investigate the sequence ͕an ͖ defined by the recurrence relation a1 ෇ 2 a nϩ1 ෇ 1 ͑a n ϩ 6͒ 2 for n ෇ 1, 2, 3, . . . SOLUTION We begin by computing the first several terms: a1 ෇ 2 a 2 ෇ 1 ͑2 ϩ 6͒ ෇ 4 2 a 3 ෇ 1 ͑4 ϩ 6͒ ෇ 5 2 a 4 ෇ 1 ͑5 ϩ 6͒ ෇ 5.5 2 a 5 ෇ 5.75 a 6 ෇ 5.875 a 7 ෇ 5.9375 a 8 ෇ 5.96875 a 9 ෇ 5.984375 5E-12(pp 746-755) 746 ❙❙❙❙ 1/18/06 10:52 AM Page 746 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| Mathematical induction is often used in dealing with recursive sequences. See page 59 for a discussion of the Principle of Mathematical Induction. These initial terms suggest that the sequence is increasing and the terms are approaching 6. To confirm that the sequence is increasing we use mathematical induction to show that a nϩ1 Ͼ a n for all n ജ 1. This is true for n ෇ 1 because a 2 ෇ 4 Ͼ a 1. If we assume that it is true for n ෇ k, then we have a kϩ1 Ͼ a k a kϩ1 ϩ 6 Ͼ a k ϩ 6 so 1 2 and ͑a kϩ1 ϩ 6͒ Ͼ 1 ͑a k ϩ 6͒ 2 a kϩ2 Ͼ a kϩ1 Thus We have deduced that a nϩ1 Ͼ a n is true for n ෇ k ϩ 1. Therefore, the inequality is true for all n by induction. Next we verify that ͕a n ͖ is bounded by showing that a n Ͻ 6 for all n. (Since the sequence is increasing, we already know that it has a lower bound: a n ജ a 1 ෇ 2 for all n.) We know that a 1 Ͻ 6, so the assertion is true for n ෇ 1. Suppose it is true for n ෇ k. Then ak Ͻ 6 a k ϩ 6 Ͻ 12 so 1 2 ͑a k ϩ 6͒ Ͻ 1 ͑12͒ ෇ 6 2 a kϩ1 Ͻ 6 Thus This shows, by mathematical induction, that a n Ͻ 6 for all n. Since the sequence ͕an ͖ is increasing and bounded, Theorem 11 guarantees that it has a limit. The theorem doesn’t tell us what the value of the limit is. But now that we know L ෇ lim n l ϱ an exists, we can use the recurrence relation to write lim a nϩ1 ෇ lim 1 ͑a n ϩ 6͒ ෇ 1 lim a n ϩ 6 ෇ 1 ͑L ϩ 6͒ 2 2 2 nlϱ |||| A proof of this fact is requested in Exercise 52. ( nlϱ nlϱ ) Since a n l L , it follows that a nϩ1 l L , too (as n l ϱ, n ϩ 1 l ϱ, too). So we have L ෇ 1 ͑L ϩ 6͒ 2 Solving this equation for L, we get L ෇ 6, as predicted. |||| 12.1 Exercises 1. (a) What is a sequence? (b) What does it mean to say that lim n l ϱ a n ෇ 8? (c) What does it mean to say that lim n l ϱ a n ෇ ϱ? 2. (a) What is a convergent sequence? Give two examples. (b) What is a divergent sequence? Give two examples. 3–8 |||| List the first five terms of the sequence. n 3. a n ෇ 1 Ϫ ͑0.2͒ 5. a n ෇ 3͑Ϫ1͒n n! 4. a n ෇ nϩ1 3n Ϫ 1 6. ͕2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͖͒ 5E-12(pp 746-755) 1/18/06 10:52 AM Page 747 SECTION 12.1 SEQUENCES 7. a 1 ෇ 3, ■ ■ 9–14 a nϩ1 ෇ 2a n Ϫ 1 ■ ■ ■ 8. a 1 ෇ 4, ■ ■ ■ a nϩ1 ෇ ■ an an Ϫ 1 ■ ■ 45. a n ෇ 47. a n ෇ 10. 11. ͕2, 7, 12, 17, . . .͖ 13. 12. 8 {1, Ϫ 2 , 4 , Ϫ 27 , . . .} 3 9 ■ ■ ■ ■ {1 , 1 , 1 , 1 , . . . } 2 4 6 8 3 4 {Ϫ 1 , 2 , Ϫ 16 , 25 , . . .} 4 9 ■ ■ ■ ■ ■ 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ n! ■ ■ |||| Determine whether the sequence converges or diverges. If it converges, find the limit. 16. a n ෇ 3 ϩ 5n 2 n ϩ n2 18. a n ෇ sn 1 ϩ sn 19. a n ෇ 2n 3 nϩ1 20. a n ෇ n 1 ϩ sn 23. a n ෇ cos͑n͞2͒ 27. ͮ ͮ e n ϩ e Ϫn e 2n Ϫ 1 1͞n 2 n arctan nlϱ 1 5n 55. a n ෇ 56. a n ෇ 2n Ϫ 3 3n ϩ 4 57. a n ෇ cos͑n␲͞2͒ 36. a n ෇ {, sin 2n 1 ϩ sn , , , , , , , . . .} ■ ■ ■ ■ 1 1 1 1 1 1 1 1 1 3 2 4 3 5 4 6 ■ ■ ■ 58. a n ෇ ne Ϫn ■ |||| Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 743 for advice on graphing sequences.) ͭ ͩ if a n is an even number if a n is an odd number 54. a n ෇ ; 41–48 43. an 3a n ϩ 1 all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit? ͑Ϫ3͒n 40. a n ෇ n! 41. a n ෇ ͑Ϫ1͒n 1 2 54–60 |||| Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? n! 39. a n ෇ n 2 ■ nϩ1 n 2n 2n ϩ 1 ͪͮ ■ 53. Suppose you know that ͕a n ͖ is a decreasing sequence and ͭ ͮ 38. ■ ■ (b) A sequence ͕a n ͖ is defined by a 1 ෇ 1 and a nϩ1 ෇ 1͑͞1 ϩ a n ͒ for n ജ 1. Assuming that ͕a n ͖ is convergent, find its limit. ln n ln 2n 37. ͕0, 1, 0, 0, 1, 0, 0, 0, 1, . . . ͖ ■ ■ 34. a n ෇ sn Ϫ sn 2 Ϫ 1 ͩ ͪ 1ϩ ■ 32. a n ෇ ln͑n ϩ 1͒ Ϫ ln n cos n 2n 33. a n ෇ n sin͑1͞n͒ 35. a n ෇ ■ lim a nϩ1 ෇ lim a n 2 31. a n ෇ ͭ nlϱ 30. ͕n cos n␲ ͖ 29. ͕n 2e Ϫn ͖ ■ 52. (a) If ͕a n ͖ is convergent, show that 26. ͕arctan 2n͖ 28. ■ 51. For what values of r is the sequence ͕nr n ͖ convergent? 24. a n ෇ cos͑2͞n͒ ͑2n Ϫ 1 ͒! ͑2n ϩ 1͒! ■ and a1 ෇ 11. Do the same if a1 ෇ 25. Make a conjecture about this type of sequence. ͑Ϫ1͒ n n3 22. a n ෇ 3 n ϩ 2n2 ϩ 1 ͭ ͭ ■ a nϩ1 ෇ ͑Ϫ1͒ nϪ1n 21. a n ෇ 2 n ϩ1 25. ■ 50. Find the first 40 terms of the sequence defined by nϩ1 3n Ϫ 1 17. a n ෇ ■ after n years the investment is worth a n ෇ 1000͑1.06͒n dollars. (a) Find the first five terms of the sequence ͕a n ͖. (b) Is the sequence convergent or divergent? Explain. ■ 15–40 15. a n ෇ n͑n Ϫ 1͒ n 46. a n ෇ s3 n ϩ 5 n 49. If $1000 is invested at 6% interest, compounded annually, then 14. ͕5, 1, 5, 1, 5, 1, . . .͖ ■ 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ ͑2n͒n 48. a n ෇ |||| 1 {1 , 1 , 1 , 16 , . . .} 2 4 8 747 ■ Find a formula for the general term a n of the sequence, assuming that the pattern of the first few terms continues. 9. n3 n! ❙❙❙❙ 42. a n ෇ 2 ϩ ͑Ϫ2͞␲ ͒n 44. ͭ ͮ sin n sn ■ 60. a n ෇ n ϩ ■ ■ ■ 59. a n ෇ 1 2n ϩ 3 n n2 ϩ 1 1 n ■ ■ ■ ■ ■ ■ ■ ■ 61. Find the limit of the sequence {s2, s2s2, s2s2s2, . . .} 62. A sequence ͕a n ͖ is given by a 1 ෇ s2, a nϩ1 ෇ s2 ϩ a n . (a) By induction or otherwise, show that ͕a n ͖ is increasing and bounded above by 3. Apply Theorem 11 to show that lim n l ϱ a n exists. (b) Find lim n l ϱ a n. ■ 5E-12(pp 746-755) 748 ❙❙❙❙ 1/18/06 10:52 AM Page 748 CHAPTER 12 INFINITE SEQUENCES AND SERIES 63. Show that the sequence defined by a 1 ෇ 1, a nϩ1 ෇ 3 Ϫ 1͞a n is increasing and a n Ͻ 3 for all n. Deduce that ͕a n ͖ is convergent and find its limit. 64. Show that the sequence defined by a1 ෇ 2 a nϩ1 ෇ (f) Use Theorem 11 to show that lim n l ϱ ͑1 ϩ 1͞n͒n exists. (The limit is e. See Equation 7.4.9.) 71. Let a and b be positive numbers with a Ͼ b. Let a 1 be their arithmetic mean and b1 their geometric mean: 1 3 Ϫ an a1 ෇ satisfies 0 Ͻ a n ഛ 2 and is decreasing. Deduce that the sequence is convergent and find its limit. 65. (a) Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? Show that the answer is fn , where ͕ fn ͖ is the Fibonacci sequence defined in Example 3(c). (b) Let a n ෇ fnϩ1͞fn and show that a nϪ1 ෇ 1 ϩ 1͞a nϪ2. Assuming that ͕a n ͖ is convergent, find its limit. 66. (a) Let a 1 ෇ a, a 2 ෇ f ͑a͒, a 3 ෇ f ͑a 2 ͒ ෇ f ͑ f ͑a͒͒, . . . , a nϩ1 ෇ f ͑a n ͒, where f is a continuous function. If lim n l ϱ a n ෇ L, show that f ͑L͒ ෇ L. (b) Illustrate part (a) by taking f ͑x͒ ෇ cos x, a ෇ 1, and estimating the value of L to five decimal places. nlϱ a nϩ1 ෇ a n ϩ bn 2 bnϩ1 ෇ sa n bn (a) Use mathematical induction to show that a n Ͼ a nϩ1 Ͼ bnϩ1 Ͼ bn (b) Deduce that both ͕a n ͖ and ͕bn ͖ are convergent. (c) Show that lim n l ϱ a n ෇ lim n l ϱ bn . Gauss called the common value of these limits the arithmetic-geometric mean of the numbers a and b. 72. (a) Show that if lim n l ϱ a 2n ෇ L and lim n l ϱ a2nϩ1 ෇ L, then ͕a n ͖ is convergent and lim n l ϱ a n ෇ L. (b) If a 1 ෇ 1 and anϩ1 ෇ 1 ϩ 1 1 ϩ an find the first eight terms of the sequence ͕a n ͖. Then use part (a) to show that lim n l ϱ a n ෇ s2. This gives the continued fraction expansion n5 n! (b) Use a graph of the sequence in part (a) to find the smallest values of N that correspond to ␧ ෇ 0.1 and ␧ ෇ 0.001 in Definition 2. 2ϩ 68. Use Definition 2 directly to prove that lim n l ϱ r ෇ 0 Խ Խ when r Ͻ 1. 1 s2 ෇ 1 ϩ n 1 2 ϩ иии 73. The size of an undisturbed fish population has been modeled by the formula 69. Prove Theorem 6. [Hint: Use either Definition 2 or the Squeeze Theorem.] ͩ ͪ n 1 . n (a) Show that if 0 ഛ a Ͻ b, then 70. Let a n ෇ b1 ෇ sab Repeat this process so that, in general, ; 67. (a) Use a graph to guess the value of the limit lim aϩb 2 1ϩ b nϩ1 Ϫ a nϩ1 Ͻ ͑n ϩ 1͒b n bϪa (b) Deduce that b n ͓͑n ϩ 1͒a Ϫ nb͔ Ͻ a nϩ1. (c) Use a ෇ 1 ϩ 1͑͞n ϩ 1͒ and b ෇ 1 ϩ 1͞n in part (b) to show that ͕a n ͖ is increasing. (d) Use a ෇ 1 and b ෇ 1 ϩ 1͑͞2n͒ in part (b) to show that a 2n Ͻ 4. (e) Use parts (c) and (d) to show that a n Ͻ 4 for all n. pnϩ1 ෇ bpn a ϩ pn where pn is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is p 0 Ͼ 0. (a) Show that if ͕ pn ͖ is convergent, then the only possible values for its limit are 0 and b Ϫ a. (b) Show that pnϩ1 Ͻ ͑ b͞a͒ pn. (c) Use part (b) to show that if a Ͼ b, then lim n l ϱ pn ෇ 0; in other words, the population dies out. (d) Now assume that a Ͻ b. Show that if p 0 Ͻ b Ϫ a, then ͕ pn ͖ is increasing and 0 Ͻ pn Ͻ b Ϫ a. Show also that if p 0 Ͼ b Ϫ a, then ͕ pn ͖ is decreasing and pn Ͼ b Ϫ a. Deduce that if a Ͻ b, then lim n l ϱ pn ෇ b Ϫ a. 5E-12(pp 746-755) 1/18/06 10:52 AM Page 749 SECTION 12.2 SERIES ❙❙❙❙ 749 LABORATORY PROJECT CAS Logistic Sequences A sequence that arises in ecology as a model for population growth is defined by the logistic difference equation pnϩ1 ෇ kpn͑1 Ϫ pn ͒ where pn measures the size of the population of the nth generation of a single species. To keep the numbers manageable, pn is a fraction of the maximal size of the population, so 0 ഛ pn ഛ 1. Notice that the form of this equation is similar to the logistic differential equation in Section 10.5. The discrete model—with sequences instead of continuous functions—is preferable for modeling insect populations, where mating and death occur in a periodic fashion. An ecologist is interested in predicting the size of the population as time goes on, and asks these questions: Will it stabilize at a limiting value? Will it change in a cyclical fashion? Or will it exhibit random behavior? Write a program to compute the first n terms of this sequence starting with an initial population p0 , where 0 Ͻ p0 Ͻ 1. Use this program to do the following. 1. Calculate 20 or 30 terms of the sequence for p0 ෇ 2 and for two values of k such that 1 1 Ͻ k Ͻ 3. Graph the sequences. Do they appear to converge? Repeat for a different value of p0 between 0 and 1. Does the limit depend on the choice of p0? Does it depend on the choice of k ? 2. Calculate terms of the sequence for a value of k between 3 and 3.4 and plot them. What do you notice about the behavior of the terms? 3. Experiment with values of k between 3.4 and 3.5. What happens to the terms? 4. For values of k between 3.6 and 4, compute and plot at least 100 terms and comment on the behavior of the sequence. What happens if you change p0 by 0.001? This type of behavior is called chaotic and is exhibited by insect populations under certain conditions. |||| 12.2 Series If we try to add the terms of an infinite sequence ͕a n ͖ϱ we get an expression of the form n෇1 1 a1 ϩ a2 ϩ a3 ϩ и и и ϩ an ϩ и и и which is called an infinite series (or just a series) and is denoted, for short, by the symbol ϱ ͚a n෇1 n or ͚a n But does it make sense to talk about the sum of infinitely many terms? It would be impossible to find a finite sum for the series 1 ϩ 2 ϩ 3 ϩ 4 ϩ 5 ϩ иии ϩ n ϩ иии because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the n th term, we get n͑n ϩ 1͒͞2, which becomes very large as n increases. However, if we start to add the terms of the series 1 1 1 1 1 1 1 ϩ ϩ ϩ ϩ ϩ ϩ иии ϩ n ϩ иии 2 4 8 16 32 64 2 5E-12(pp 746-755) ❙❙❙❙ 750 1/18/06 10:52 AM Page 750 CHAPTER 12 INFINITE SEQUENCES AND SERIES n Sum of first n terms 1 2 3 4 5 6 7 10 15 20 25 0.50000000 0.75000000 0.87500000 0.93750000 0.96875000 0.98437500 0.99218750 0.99902344 0.99996948 0.99999905 0.99999997 we get 1 , 3 , 7 , 15 , 31 , 63 , . . . , 1 Ϫ 1͞2 n, . . . . The table shows that as we add more and more 2 4 8 16 32 64 terms, these partial sums become closer and closer to 1. (See also Figure 11 in A Preview of Calculus, page 7.) In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write ϱ ͚ n෇1 1 1 1 1 1 1 ෇ ϩ ϩ ϩ ϩ иии ϩ n ϩ иии ෇ 1 2n 2 4 8 16 2 We use a similar idea to determine whether or not a general series (1) has a sum. We consider the partial sums s1 ෇ a 1 s2 ෇ a 1 ϩ a 2 s3 ෇ a 1 ϩ a 2 ϩ a 3 s4 ෇ a 1 ϩ a 2 ϩ a 3 ϩ a 4 and, in general, n sn ෇ a 1 ϩ a 2 ϩ a 3 ϩ и и и ϩ a n ෇ ͚a i i෇1 These partial sums form a new sequence ͕sn ͖, which may or may not have a limit. If lim n l ϱ sn ෇ s exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series ͸ a n . 2 Definition Given a series ͸ϱ a n ෇ a 1 ϩ a 2 ϩ a 3 ϩ и и и , let sn denote its nth n෇1 partial sum: n sn ෇ ͚a i ෇ a1 ϩ a2 ϩ и и и ϩ an i෇1 If the sequence ͕sn ͖ is convergent and lim n l ϱ sn ෇ s exists as a real number, then the series ͸ a n is called convergent and we write ϱ a1 ϩ a2 ϩ и и и ϩ an ϩ и и и ෇ s ͚a or n ෇s n෇1 The number s is called the sum of the series. Otherwise, the series is called divergent. Thus, when we write ͸ϱ a n ෇ s we mean that by adding sufficiently many terms of n෇1 the series we can get as close as we like to the number s. Notice that |||| Compare with the improper integral y ϱ 1 f ͑x͒ dx ෇ lim tlϱ y t 1 f ͑x͒ dx To find this integral we integrate from 1 to t and then let t l ϱ. For a series, we sum from 1 to n and then let n l ϱ. ϱ ͚a n෇1 n n ෇ lim ͚a n l ϱ i෇1 i EXAMPLE 1 An important example of an infinite series is the geometric series a ϩ ar ϩ ar 2 ϩ ar 3 ϩ и и и ϩ ar nϪ1 ϩ и и и ෇ ϱ ͚ ar n෇1 nϪ1 a 0 5E-12(pp 746-755) 1/18/06 10:52 AM Page 751 SECTION 12.2 SERIES |||| Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles, s a ෇ a a Ϫ ar so s෇ a 1Ϫr sn ෇ a ϩ ar ϩ ar 2 ϩ и и и ϩ ar nϪ1 rsn ෇ ar ϩ ar 2 ϩ и и и ϩ ar nϪ1 ϩ ar n and ar@ Subtracting these equations, we get ar@ sn Ϫ rsn ෇ a Ϫ ar n ar ar a 751 Each term is obtained from the preceding one by multiplying it by the common ratio r. (We have already considered the special case where a ෇ 1 and r ෇ 1 on page 749.) 2 2 If r ෇ 1, then sn ෇ a ϩ a ϩ и и и ϩ a ෇ na l Ϯϱ. Since lim n l ϱ sn doesn’t exist, the geometric series diverges in this case. If r 1, we have ar# a-ar ❙❙❙❙ s a sn ෇ 3 If Ϫ1 Ͻ r Ͻ 1, we know from (12.1.8) that r n l 0 as n l ϱ, so lim sn ෇ lim nlϱ a FIGURE 1 a͑1 Ϫ r n ͒ 1Ϫr nlϱ a͑1 Ϫ r n ͒ a a a ෇ Ϫ lim r n ෇ 1Ϫr 1Ϫr 1 Ϫ r nlϱ 1Ϫr Խ Խ Thus, when r Ͻ 1 the geometric series is convergent and its sum is a͑͞1 Ϫ r͒. If r ഛ Ϫ1 or r Ͼ 1, the sequence ͕r n ͖ is divergent by (12.1.8) and so, by Equation 3, lim n l ϱ sn does not exist. Therefore, the geometric series diverges in those cases. We summarize the results of Example 1 as follows. 4 The geometric series ϱ ͚ ar nϪ1 ෇ a ϩ ar ϩ ar 2 ϩ и и и n෇1 |||| In words: The sum of a convergent geometric series is Խ Խ is convergent if r Ͻ 1 and its sum is ϱ ͚ ar first term 1 Ϫ common ratio nϪ1 ෇ n෇1 a 1Ϫr ԽrԽ Ͻ 1 Խ Խ If r ജ 1, the geometric series is divergent. EXAMPLE 2 Find the sum of the geometric series 5 Ϫ 10 ϩ 20 Ϫ 40 ϩ и и и 3 9 27 Խ Խ SOLUTION The first term is a ෇ 5 and the common ratio is r ෇ Ϫ3 . Since r ෇ 2 series is convergent by (4) and its sum is 5Ϫ 10 20 40 5 5 ϩ Ϫ ϩ иии ෇ ෇ 5 ෇3 3 9 27 1 Ϫ (Ϫ 2 ) 3 3 2 3 Ͻ 1, the 5E-12(pp 746-755) 752 ❙❙❙❙ 1/18/06 10:52 AM Page 752 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we can’t literally add an infinite number of terms, one by one. But, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3. The table shows the first ten partial sums sn and the graph in Figure 2 shows how the sequence of partial sums approaches 3. sn n sn 1 2 3 4 5 6 7 8 9 10 5.000000 1.666667 3.888889 2.407407 3.395062 2.736626 3.175583 2.882945 3.078037 2.947975 3 0 20 n FIGURE 2 ϱ EXAMPLE 3 Is the series ͚2 2n 3 1Ϫn convergent or divergent? n෇1 SOLUTION Let’s rewrite the nth term of the series in the form ar nϪ1: |||| Another way to identify a and r is to write out the first few terms: 4ϩ 16 3 ϱ ͚2 ϩ 64 ϩ и и и 9 2n 3 1Ϫn ෇ n෇1 ϱ ͚ n෇1 ϱ 4n nϪ1 ෇ ͚ 4( 4 ) 3 3 nϪ1 n෇1 We recognize this series as a geometric series with a ෇ 4 and r ෇ 4 . Since r Ͼ 1, the 3 series diverges by (4). EXAMPLE 4 Write the number 2.317 ෇ 2.3171717. . . as a ratio of integers. SOLUTION 17 17 17 ϩ 5 ϩ 7 ϩ иии 10 3 10 10 2.3171717. . . ෇ 2.3 ϩ After the first term we have a geometric series with a ෇ 17͞10 3 and r ෇ 1͞10 2. Therefore 17 17 3 10 1000 2.317 ෇ 2.3 ϩ ෇ 2.3 ϩ 1 99 1Ϫ 2 10 100 23 17 1147 ෇ ϩ ෇ 10 990 495 ϱ EXAMPLE 5 Find the sum of the series ͚ x , where Խ x Խ Ͻ 1. n n෇0 SOLUTION Notice that this series starts with n ෇ 0 and so the first term is x 0 ෇ 1. (With Module 12.2 explores a series that depends on an angle ␪ in a triangle and enables you to see how rapidly the series converges when ␪ varies. series, we adopt the convention that x 0 ෇ 1 even when x ෇ 0.) Thus ϱ ͚x n ෇ 1 ϩ x ϩ x2 ϩ x3 ϩ x4 ϩ и и и n෇0 Խ Խ Խ Խ This is a geometric series with a ෇ 1 and r ෇ x. Since r ෇ x Ͻ 1, it converges and (4) gives ϱ 5 ͚x n෇0 n ෇ 1 1Ϫx 5E-12(pp 746-755) 1/18/06 10:53 AM Page 753 SECTION 12.2 SERIES ϱ ͚ EXAMPLE 6 Show that the series n෇1 ❙❙❙❙ 753 1 is convergent, and find its sum. n͑n ϩ 1͒ SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums. n sn ෇ ͚ i෇1 1 1 1 1 1 ෇ ϩ ϩ ϩ иии ϩ i͑i ϩ 1͒ 1ؒ2 2ؒ3 3ؒ4 n͑n ϩ 1͒ We can simplify this expression if we use the partial fraction decomposition 1 1 1 ෇ Ϫ i͑i ϩ 1͒ i iϩ1 (see Section 8.4). Thus, we have n sn ෇ ͚ i෇1 n 1 ෇͚ i͑i ϩ 1͒ i෇1 ෇1Ϫ 1 2 1 1 Ϫ 2 3 ϩ 1 nϩ1 lim sn ෇ lim and so nlϱ ͪ nlϱ 1 1 Ϫ 3 4 ϩ ͩ 1Ϫ 1 nϩ1 ͪ ϩ иии ϩ ͩ 1 1 Ϫ n nϩ1 ෇1Ϫ0෇1 Therefore, the given series is convergent and ϱ ͚ n෇1 1 ෇1 n͑n ϩ 1͒ EXAMPLE 7 Show that the harmonic series ϱ ͚ 1 n෇1 ͕sn ͖ 1 1 Ϫ i iϩ1 ͩ ͪ ͩ ͪ ͩ ͪ ෇ 1Ϫ |||| Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like an old-fashioned collapsing telescope) into just two terms. |||| Figure 3 illustrates Example 6 by showing the graphs of the sequence of terms a n ෇ 1͞[n͑n ϩ 1͒] and the sequence ͕sn ͖ of partial sums. Notice that a n l 0 and sn l 1. See Exercises 54 and 55 for two geometric interpretations of Example 6. ͩ 1 1 1 1 ෇ 1 ϩ ϩ ϩ ϩ иии n 2 3 4 is divergent. SOLUTION s1 ෇ 1 s2 ෇ 1 ϩ 1 2 ͕an ͖ 0 FIGURE 3 n s4 ෇ 1 ϩ 1 ϩ ( 1 ϩ 1 ) Ͼ 1 ϩ 1 ϩ ( 1 ϩ 1 ) ෇ 1 ϩ 2 2 3 4 2 4 4 2 s8 ෇ 1 ϩ 1 ϩ ( 1 ϩ 1 ) ϩ ( 1 ϩ 1 ϩ 1 ϩ 1 ) 2 3 4 5 6 7 8 Ͼ 1 ϩ 1 ϩ (1 ϩ 1) ϩ (1 ϩ 1 ϩ 1 ϩ 1) 2 4 4 8 8 8 8 ෇1ϩ1ϩ1ϩ1෇1ϩ3 2 2 2 2 1 s16 ෇ 1 ϩ 1 ϩ ( 1 ϩ 1 ) ϩ ( 1 ϩ и и и ϩ 1 ) ϩ ( 1 ϩ и и и ϩ 16 ) 2 3 4 5 8 9 1 1 Ͼ 1 ϩ 1 ϩ ( 1 ϩ 1 ) ϩ ( 1 ϩ и и и ϩ 1 ) ϩ ( 16 ϩ и и и ϩ 16 ) 2 4 4 8 8 ෇1ϩ1ϩ1ϩ1ϩ1෇1ϩ4 2 2 2 2 2 ͪ 5E-12(pp 746-755) 754 ❙❙❙❙ 1/18/06 10:53 AM Page 754 CHAPTER 12 INFINITE SEQUENCES AND SERIES Similarly, s32 Ͼ 1 ϩ 5 , s64 Ͼ 1 ϩ 6 , and in general 2 2 s2 n Ͼ 1 ϩ |||| The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382). n 2 This shows that s2 n l ϱ as n l ϱ and so ͕sn ͖ is divergent. Therefore, the harmonic series diverges. ϱ 6 Theorem If the series ͚a n is convergent, then lim an ෇ 0. nlϱ n෇1 ͸ a n is convergent, the sequence ͕sn ͖ is convergent. Let lim n l ϱ sn ෇ s. Since n Ϫ 1 l ϱ as n l ϱ, we also have lim n l ϱ snϪ1 ෇ s. Therefore Proof Let sn ෇ a 1 ϩ a 2 ϩ и и и ϩ a n . Then a n ෇ sn Ϫ snϪ1. Since lim a n ෇ lim ͑sn Ϫ snϪ1 ͒ ෇ lim sn Ϫ lim snϪ1 nlϱ nlϱ nlϱ nlϱ ෇sϪs෇0 With any series ͸ a n we associate two sequences: the sequence ͕sn ͖ of its partial sums and the sequence ͕a n ͖ of its terms. If ͸ a n is convergent, then the limit of the sequence ͕sn ͖ is s (the sum of the series) and, as Theorem 6 asserts, the limit of the sequence ͕a n ͖ is 0. NOTE 1 | ■ The converse of Theorem 6 is not true in general. If lim n l ϱ a n ෇ 0, we cannot conclude that ͸ a n is convergent. Observe that for the harmonic series ͸ 1͞n we have a n ෇ 1͞n l 0 as n l ϱ, but we showed in Example 7 that ͸ 1͞n is divergent. NOTE 2 7 ■ The Test for Divergence If lim a n does not exist or if lim a n nlϱ ϱ series ͚a n nlϱ 0, then the is divergent. n෇1 The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so lim n l ϱ a n ෇ 0. ϱ EXAMPLE 8 Show that the series ͚ n෇1 n2 diverges. 5n 2 ϩ 4 SOLUTION lim a n ෇ lim nlϱ nlϱ n2 1 1 ෇ lim ෇ 2 n l ϱ 5 ϩ 4͞n 2 5n ϩ 4 5 0 So the series diverges by the Test for Divergence. NOTE 3 If we find that lim n l ϱ a n 0, we know that ͸ a n is divergent. If we find that lim n l ϱ a n ෇ 0, we know nothing about the convergence or divergence of ͸ a n. Remember the warning in Note 2: If lim n l ϱ a n ෇ 0, the series ͸ a n might converge or it might diverge. ■ 5E-12(pp 746-755) 1/18/06 10:53 AM Page 755 SECTION 12.2 SERIES ❙❙❙❙ 755 8 Theorem If ͸ a n and ͸ bn are convergent series, then so are the series ͸ ca n (where c is a constant), ͸ ͑a n ϩ bn ͒, and ͸ ͑a n Ϫ bn ͒, and ϱ (i) ͚ ca n ෇c n෇1 ͚ ͑a ϱ ͚ ͑a (ii) n n෇1 ϱ (iii) ϱ ͚a n Ϫ bn ͒ ෇ n෇1 n ϱ ͚a ϩ bn ͒ ෇ n෇1 ϱ ͚a n Ϫ n෇1 n ϩ n෇1 ϱ ͚b n n෇1 ϱ ͚b n n෇1 These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1. For instance, here is how part (ii) of Theorem 8 is proved: Let ϱ n ͚a sn ෇ ͚a s෇ i i෇1 n ͚b tn ෇ n t෇ i n෇1 i෇1 ϱ ͚b n n෇1 The nth partial sum for the series ͸ ͑a n ϩ bn ͒ is n un ෇ ͚ ͑a i ϩ bi ͒ i෇1 and, using Equation 5.2.9, we have n lim u n ෇ lim nlϱ i n l ϱ i෇1 ϩ bi ͒ ෇ lim nlϱ n ෇ lim n ͚b ai ϩ i෇1 ͪ i i෇1 n ͚a i n l ϱ i෇1 ͚ͩ n ͚ ͑a ϩ lim ͚b i n l ϱ i෇1 ෇ lim sn ϩ lim tn ෇ s ϩ t nlϱ nlϱ Therefore, ͸ ͑a n ϩ bn ͒ is convergent and its sum is ϱ ͚ ͑a n n෇1 ͚a n n෇1 ϱ ͚ EXAMPLE 9 Find the sum of the series n෇1 SOLUTION The series ϱ ϩ bn ͒ ෇ s ϩ t ෇ ͩ ϩ ϱ ͚b n n෇1 ͪ 3 1 ϩ n . n͑n ϩ 1͒ 2 ͸ 1͞2 n is a geometric series with a ෇ 1 and r ෇ 1 , so 2 2 ϱ ͚ n෇1 1 1 ෇ 2 1 ෇1 2n 1Ϫ2 In Example 6 we found that ϱ ͚ n෇1 1 ෇1 n͑n ϩ 1͒ So, by Theorem 8, the given series is convergent and ϱ ͚ n෇1 ͩ 3 1 ϩ n n͑n ϩ 1͒ 2 ͪ ෇3 ϱ ͚ n෇1 ϱ 1 1 ϩ ͚ n n͑n ϩ 1͒ n෇1 2 ෇3ؒ1ϩ1෇4 5E-12(pp 756-765) 756 ❙❙❙❙ 1/18/06 10:09 AM Page 756 CHAPTER 12 INFINITE SEQUENCES AND SERIES NOTE 4 A finite number of terms doesn’t affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series ■ ϱ ͚ n෇4 n n3 ϩ 1 is convergent. Since ϱ ͚ n෇1 ϱ n 1 2 3 n ෇ ϩ ϩ ϩ ͚ 3 3 n ϩ1 2 9 28 n෇4 n ϩ 1 it follows that the entire series ͸ϱ n͑͞n 3 ϩ 1͒ is convergent. Similarly, if it is known that n෇1 the series ͸ϱ n෇Nϩ1 a n converges, then the full series ϱ ϱ N ͚a n ෇ n෇1 ͚a n ͚ ϩ n෇1 an n෇Nϩ1 is also convergent. |||| 12.2 Exercises 14. 1 ϩ 0.4 ϩ 0.16 ϩ 0.064 ϩ и и и 1. (a) What is the difference between a sequence and a series? (b) What is a convergent series? What is a divergent series? ϱ 2. Explain what it means to say that ͸ϱ a n ෇ 5. n෇1 15. ϱ |||| Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why. ϱ ͚ n෇1 ϱ 12 ͑Ϫ5͒n 4. ϱ 5. ϱ 7. ͚ n෇1 ■ ͩ 6. 1 1 Ϫ n 1.5 ͑n ϩ 1͒1.5 ■ 17. ϱ 19. ϱ 21. ■ ■ ■ ϱ 8. ͚ n෇2 ■ ■ ϱ 23. ■ ϱ ■ ■ 2n . 3n ϩ 1 (a) Determine whether ͕a n ͖ is convergent. (b) Determine whether ͸ϱ a n is convergent. n෇1 ■ 25. ͚ ͚ i෇1 ϱ 27. ͚a 31. ͚ n෇1 13. Ϫ2 ϩ 2 Ϫ 5 8 25 8 ϩ 125 Ϫ и и и 32 22. 2 2 n Ϫ1 24. n ͚ n෇1 ϱ ͚ n෇1 en 3 nϪ1 3 n ϱ n෇1 ϱ 2 k k Ϫ1 26. 3n ϩ 2n 6n 28. 2 j ■ ͩ ■ ͚ ͚ n෇1 ͑n ϩ 1͒2 n͑n ϩ 2͒ 2 n 2 ϩ 4n ϩ 3 ϱ ͚ ͓͑0.8͒ n෇1 ͩ ϱ 30. ͚ ln n෇1 Ϫ ͑0.3͒ n ͔ n 2n ϩ 5 ͪ ϱ 32. 3 5 ϩ n n͑n ϩ 3͒ 4 ■ nϪ1 ■ k෇1 ͪ ■ ͚ ͑cos 1͒ ϱ 34. ͚ n෇1 ■ ■ ͩ k 3 2 ϩ 5n n ■ ■ ͪ ■ ■ ■ i෇1 Determine whether the series is convergent or divergent. If it is convergent, find its sum. 4 n nϩ5 ϱ ͚ arctan n ϱ |||| 11. 3 ϩ 2 ϩ 3 ϩ 9 ϩ и и и 20. 1 ͚ (s2 ) n෇0 n 35–40 11–34 ␲n 3 nϩ1 ͑Ϫ6͒ nϪ1 5 nϪ1 ϱ ͚ s2 n෇1 33. ͚a and 18. ϱ 29. n i ͑Ϫ3͒ nϪ1 4n ϱ aj (b) Explain the difference between i෇1 ͚ n෇1 j෇1 n ͚ n෇1 n෇1 n and ͚ k෇2 10. (a) Explain the difference between ai ͚ n෇2 9. Let a n ෇ n 16. nϪ1 1 n͑n Ϫ 1͒ ■ ͚ n෇1 n෇1 ͪ ͚ n෇0 2n 2 Ϫ 1 n2 ϩ 1 ͚ ͑0.6͒ ͚ n෇1 ϱ ͚ tan n n෇1 ͚ n෇1 2 nϪ1 3 n෇1 ; 3–8 3. ϱ ͚ 5( ) 12. 1 8 Ϫ 1 ϩ 1 Ϫ 1 ϩ иии 4 2 |||| Express the number as a ratio of integers. 35. 0.2 ෇ 0.2222 . . . 36. 0.73 ෇ 0.73737373 . . . 37. 3.417 ෇ 3.417417417 . . . 38. 6.254 39. 0.123456 ■ ■ ■ 40. 5.6021 ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-12(pp 756-765) 1/18/06 10:09 AM Page 757 ❙❙❙❙ SECTION 12.2 SERIES 41–45 |||| Find the values of x for which the series converges. Find the sum of the series for those values of x. ϱ 41. ͚ n෇1 ͚ ͑x Ϫ 4͒ 42. ͚4x n ϱ n 45. ͚ n෇0 ■ ͚ 44. n෇0 ϱ n n෇1 ϱ 43. (a) Assuming that the ball continues to bounce indefinitely, find the total distance that it travels. (Use the fact that the ball falls 1 tt 2 meters in t seconds.) 2 (b) Calculate the total time that the ball travels. (c) Suppose that each time the ball strikes the surface with velocity v it rebounds with velocity Ϫk v, where 0 Ͻ k Ͻ 1. How long will it take for the ball to come to rest? ϱ xn 3n n෇0 ͑x ϩ 3͒ n 2n n cos x 2n ■ ϱ 53. What is the value of c if ■ ■ ■ ■ ■ ■ ■ ■ ■ n෇1 1 n ϱ ͚ is another series with this property. n෇1 CAS 47–48 Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using the CAS to sum the series directly. |||| ϱ 47. ͚ n෇1 ■ ■ ϱ 1 ͑4n ϩ 1͒͑4n Ϫ 3͒ ■ ■ ■ 48. ■ 49. If the nth partial sum of a series ͸ sn ෇ n 2 ϩ 3n ϩ 1 ͑n 2 ϩ n͒2 ■ ϱ n෇1 find a n and ͸ϱ an . n෇1 ͚ n෇1 ■ ■ ■ ■ ■ 1 ෇1 n͑n ϩ 1͒ 55. The figure shows two circles C and D of radius 1 that touch at P. T is a common tangent line; C1 is the circle that touches C, D, and T ; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2. This procedure can be continued indefinitely and produces an infinite sequence of circles ͕Cn ͖. Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 6. an is nϪ1 nϩ1 50. If the nth partial sum of a series ͸ϱ an is sn ෇ 3 Ϫ n 2 Ϫn, n෇1 find a n and ͸ ϱ n෇1 ෇ 2? on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that ͩ ͪ 1ϩ Ϫn n ; 54. Graph the curves y ෇ x , 0 ഛ x ഛ 1, for n ෇ 0, 1, 2, 3, 4, . . . whose terms approach 0. Show that ϱ ͚ ͑1 ϩ c͒ n෇2 ■ 46. We have seen that the harmonic series is a divergent series ͚ ln 757 P an . 51. When money is spent on goods and services, those that receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending D dollars. Suppose that each recipient of spent money spends 100c% and saves 100s% of the money that he or she receives. The values c and s are called the marginal propensity to consume and the marginal propensity to save and, of course, c ϩ s ෇ 1. (a) Let Sn be the total spending that has been generated after n transactions. Find an equation for Sn. (b) Show that lim n l ϱ Sn ෇ kD, where k ෇ 1͞s. The number k is called the multiplier. What is the multiplier if the marginal propensity to consume is 80%? Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits. C£ C™ 1 C 1 D C¡ T Խ CD is drawn perpendicular to AB, DE is drawn perpendicular to BC, EF Ќ AB, and this process is continued indefinitely as shown in the figure. Find the total length of all the perpendiculars Խ CD Խ ϩ Խ DE Խ ϩ Խ EF Խ ϩ Խ FG Խ ϩ и и и in terms of b and ␪. A D ¨ F H b 52. A certain ball has the property that each time it falls from a height h onto a hard, level surface, it rebounds to a height rh, where 0 Ͻ r Ͻ 1. Suppose that the ball is dropped from an initial height of H meters. Խ 56. A right triangle ABC is given with ЄA ෇ ␪ and AC ෇ b. B G E C 5E-12(pp 756-765) 758 ❙❙❙❙ 1/18/06 10:10 AM Page 758 CHAPTER 12 INFINITE SEQUENCES AND SERIES 57. What is wrong with the following calculation? 0 ෇ 0 ϩ 0 ϩ 0 ϩ иии ෇ ͑1 Ϫ 1͒ ϩ ͑1 Ϫ 1͒ ϩ ͑1 Ϫ 1͒ ϩ и и и ෇ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ иии ෇ 1 ϩ ͑Ϫ1 ϩ 1͒ ϩ ͑Ϫ1 ϩ 1͒ ϩ ͑Ϫ1 ϩ 1͒ ϩ и и и ෇ 1 ϩ 0 ϩ 0 ϩ 0 ϩ иии ෇ 1 nitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center oneninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0. (Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”) 58. Suppose that ͸ϱ a n ͑a n n෇1 0͒ is known to be a convergent series. Prove that ͸ϱ 1͞a n is a divergent series. n෇1 59. Prove part (i) of Theorem 8. 60. If ͸ a n is divergent and c 0, show that ͸ ca n is divergent. 61. If ͸ a n is convergent and ͸ bn is divergent, show that the series 62. ͸ ͑a n ϩ bn ͒ is divergent. [Hint: Argue by contradiction.] If ͸ a n and ͸ bn are both divergent, is ͸ ͑a n ϩ bn ͒ necessarily divergent? 63. Suppose that a series ͸ a n has positive terms and its partial sums sn satisfy the inequality sn ഛ 1000 for all n. Explain why ͸ a n must be convergent. 64. The Fibonacci sequence was defined in Section 12.1 by the equations f1 ෇ 1, f2 ෇ 1, fn ෇ fnϪ1 ϩ fnϪ2 nജ3 Show that each of the following statements is true. 1 1 1 (a) ෇ Ϫ fnϪ1 fnϩ1 fnϪ1 fn fn fnϩ1 ϱ ͚ 1 ෇1 fnϪ1 fnϩ1 ϱ (b) fn ෇2 fnϪ1 fnϩ1 n෇2 (c) ͚ n෇2 65. The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval [0, 1] and remove the open interval ( 1 , 2 ). 3 3 That leaves the two intervals [0, 1 ] and [ 2, 1] and we remove 3 3 the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in [0, 1] after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains infi- 66. (a) A sequence ͕a n ͖ is defined recursively by the equation a n ෇ 1 ͑a nϪ1 ϩ a nϪ2 ͒ for n ജ 3, where a 1 and a 2 can be any 2 real numbers. Experiment with various values of a 1 and a 2 and use your calculator to guess the limit of the sequence. (b) Find lim n l ϱ a n in terms of a 1 and a 2 by expressing a nϩ1 Ϫ a n in terms of a 2 Ϫ a 1 and summing a series. 67. Consider the series ϱ ͚ n෇1 n ͑n ϩ 1͒! (a) Find the partial sums s1, s2, s3, and s4. Do you recognize the denominators? Use the pattern to guess a formula for sn . (b) Use mathematical induction to prove your guess. (c) Show that the given infinite series is convergent, and find its sum. 68. In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles. 5E-12(pp 756-765) 1/18/06 10:10 AM Page 759 SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS |||| 12.3 ❙❙❙❙ 759 The Integral Test and Estimates of Sums In general, it is difficult to find the exact sum of a series. We were able to accomplish this for geometric series and the series ͸ 1͓͞n͑n ϩ 1͔͒ because in each of those cases we could find a simple formula for the nth partial sum sn . But usually it isn’t easy to compute lim n l ϱ sn. Therefore, in the next few sections we develop several tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. (In some cases, however, our methods will enable us to find good estimates of the sum.) Our first test involves improper integrals. We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers: n n sn ෇ ͚ i෇1 5 10 50 100 500 1000 5000 ϱ 1 i2 1.4636 1.5498 1.6251 1.6350 1.6429 1.6439 1.6447 ͚ n෇1 1 1 1 1 1 1 ෇ 2 ϩ 2 ϩ 2 ϩ 2 ϩ 2 ϩ иии n2 1 2 3 4 5 There’s no simple formula for the sum sn of the first n terms, but the computer-generated table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as n l ϱ and so it looks as if the series is convergent. We can confirm this impression with a geometric argument. Figure 1 shows the curve y ෇ 1͞x 2 and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function y ෇ 1͞x 2 at the right endpoint of the interval. So the sum of the areas of the rectangles is ϱ 1 1 1 1 1 1 2 ϩ 2 ϩ 2 ϩ 2 ϩ 2 ϩ иии ෇ ͚ 1 2 3 4 5 n2 n෇1 y y= 1 ≈ area= 1 1@ 0 1 2 area= 1 2@ FIGURE 1 3 area= 1 3@ 4 area= 1 4@ 5 x area= 1 5@ If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y ෇ 1͞x 2 for x ജ 1, which is the value of the integral x1ϱ ͑1͞x 2 ͒ dx. In Section 8.8 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than ϱ 1 1 2 ϩ y 2 dx ෇ 2 1 x 1 Thus, the partial sums are bounded. We also know that the partial sums are increasing (because all the terms are positive). Therefore, the partial sums converge (by the Monotonic Sequence Theorem) and so the series is convergent. The sum of the series (the limit of the partial sums) is also less than 2: ϱ ͚ n෇1 1 1 1 1 1 ϩ иии Ͻ 2 2 ෇ 2 ϩ 2 ϩ 2 ϩ n 1 2 3 42 5E-12(pp 756-765) 760 ❙❙❙❙ 1/18/06 10:10 AM Page 760 CHAPTER 12 INFINITE SEQUENCES AND SERIES n n sn ෇ ͚ i෇1 5 10 50 100 500 1000 5000 1 si 3.2317 5.0210 12.7524 18.5896 43.2834 61.8010 139.9681 [The exact sum of this series was found by the Swiss mathematician Leonhard Euler (1707–1783) to be ␲ 2͞6, but the proof of this fact is quite difficult. (See Problem 6 in the Problems Plus following Chapter 16.)] Now let’s look at the series ϱ ͚ n෇1 1 1 1 1 1 1 ෇ ϩ ϩ ϩ ϩ ϩ иии sn s1 s2 s3 s4 s5 The table of values of sn suggests that the partial sums aren’t approaching a finite number, so we suspect that the given series may be divergent. Again we use a picture for confirmation. Figure 2 shows the curve y ෇ 1͞sx, but this time we use rectangles whose tops lie above the curve. y y= 1 x œ„ 0 1 2 area= 1 1 œ„ FIGURE 2 3 area= 1 2 œ„ 4 area= 1 3 œ„ 5 x area= 1 4 œ„ The base of each rectangle is an interval of length 1. The height is equal to the value of the function y ෇ 1͞sx at the left endpoint of the interval. So the sum of the areas of all the rectangles is ϱ 1 1 1 1 1 1 ϩ ϩ ϩ ϩ ϩ иии ෇ ͚ s1 s2 s3 s4 s5 n෇1 sn This total area is greater than the area under the curve y ෇ 1͞sx for x ജ 1, which is equal to the integral x1ϱ (1͞sx ) dx. But we know from Section 8.8 that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite; that is, the series is divergent. The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.) The Integral Test Suppose f is a continuous, positive, decreasing function on ͓1, ϱ͒ and let a n ෇ f ͑n͒. Then the series ͸ϱ a n is convergent if and only if the improper n෇1 integral x1ϱ f ͑x͒ dx is convergent. In other words: ϱ ϱ ͚a (i) If y f ͑x͒ dx is convergent, then 1 n is convergent. n෇1 ϱ (ii) If y f ͑x͒ dx is divergent, then 1 ϱ ͚a n is divergent. n෇1 NOTE When we use the Integral Test it is not necessary to start the series or the integral at n ෇ 1. For instance, in testing the series ■ ϱ ͚ n෇4 1 ͑n Ϫ 3͒2 we use y ϱ 4 1 dx ͑x Ϫ 3͒2 5E-12(pp 756-765) 1/18/06 10:10 AM Page 761 SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 761 Also, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. Then ͸ϱ a n is n෇N convergent, so ͸ϱ a n is convergent by Note 4 of Section 12.2. n෇1 ϱ EXAMPLE 1 Test the series ͚ n෇1 1 for convergence or divergence. n2 ϩ 1 SOLUTION The function f ͑x͒ ෇ 1͑͞x 2 ϩ 1͒ is continuous, positive, and decreasing on ͓1, ϱ͒ so we use the Integral Test: y 1 t 1 dx ෇ lim y 2 dx ෇ lim tanϪ1x tlϱ 1 x ϩ 1 tlϱ x2 ϩ 1 ϱ 1 ͩ ෇ lim tanϪ1t Ϫ tlϱ ␲ 4 ͪ ෇ t 1 ␲ ␲ ␲ Ϫ ෇ 2 4 4 Thus, x1ϱ 1͑͞x 2 ϩ 1͒ dx is a convergent integral and so, by the Integral Test, the series ͸ 1͑͞n 2 ϩ 1͒ is convergent. ϱ EXAMPLE 2 For what values of p is the series ͚ n෇1 1 convergent? np SOLUTION If p Ͻ 0, then lim n l ϱ ͑1͞n ͒ ෇ ϱ. If p ෇ 0, then lim n l ϱ ͑1͞n p ͒ ෇ 1. In either p case lim n l ϱ ͑1͞n p ͒ 0, so the given series diverges by the Test for Divergence (12.2.7). If p Ͼ 0, then the function f ͑x͒ ෇ 1͞x p is clearly continuous, positive, and decreasing on ͓1, ϱ͒. We found in Chapter 8 [see (8.8.2)] that y 1 dx converges if p Ͼ 1 and diverges if p ഛ 1 xp ϱ 1 It follows from the Integral Test that the series ͸ 1͞n p converges if p Ͼ 1 and diverges if 0 Ͻ p ഛ 1. (For p ෇ 1, this series is the harmonic series discussed in Example 7 in Section 12.2.) The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows. ϱ 1 The p-series ͚ n෇1 1 is convergent if p Ͼ 1 and divergent if p ഛ 1. np EXAMPLE 3 (a) The series ϱ ͚ n෇1 1 1 1 1 1 ෇ 3 ϩ 3 ϩ 3 ϩ 3 ϩ иии n3 1 2 3 4 is convergent because it is a p-series with p ෇ 3 Ͼ 1. (b) The series ϱ ͚ n෇1 ϱ 1 1 1 1 1 ෇ 1 ϩ 3 ϩ 3 ϩ 3 ϩ иии 1͞3 ෇ ͚ 3 n s2 s3 s4 n෇1 sn is divergent because it is a p-series with p ෇ 1 Ͻ 1. 3 5E-12(pp 756-765) 762 ❙❙❙❙ 1/18/06 10:10 AM Page 762 CHAPTER 12 INFINITE SEQUENCES AND SERIES NOTE We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact, ■ ϱ ͚ n෇1 1 ␲2 ෇ n2 6 whereas y ϱ 1 1 dx ෇ 1 x2 Therefore, in general, ϱ ͚a y n ϱ EXAMPLE 4 Determine whether the series ϱ 1 n෇1 ͚ n෇1 f ͑x͒ dx ln n converges or diverges. n SOLUTION The function f ͑x͒ ෇ ͑ln x͒͞x is positive and continuous for x Ͼ 1 because the logarithm function is continuous. But it is not obvious whether or not f is decreasing, so we compute its derivative: f Ј͑x͒ ෇ ͑1͞x͒x Ϫ ln x 1 Ϫ ln x ෇ 2 x x2 Thus, f Ј͑x͒ Ͻ 0 when ln x Ͼ 1, that is, x Ͼ e. It follows that f is decreasing when x Ͼ e and so we can apply the Integral Test: y ϱ 1 ln x t ln x ͑ln x͒2 dx ෇ lim y dx ෇ lim tlϱ 1 tlϱ x x 2 ͬ t 1 ͑ln t͒2 ෇ lim ෇ϱ tlϱ 2 Since this improper integral is divergent, the series ͸ ͑ln n͒͞n is also divergent by the Integral Test. y Estimating the Sum of a Series y=ƒ an+1 an+2 0 Suppose we have been able to use the Integral Test to show that a series ͸ a n is convergent and we now want to find an approximation to the sum s of the series. Of course, any partial sum sn is an approximation to s because lim n l ϱ sn ෇ s. But how good is such an approximation? To find out, we need to estimate the size of the remainder ... x n Rn ෇ s Ϫ sn ෇ a nϩ1 ϩ a nϩ2 ϩ a nϩ3 ϩ и и и The remainder Rn is the error made when sn, the sum of the first n terms, is used as an approximation to the total sum. We use the same notation and ideas as in the Integral Test, assuming that f is decreasing on ͓n, ϱ͒. Comparing the areas of the rectangles with the area under y ෇ f ͑x͒ for x Ͼ n in Figure 3, we see that FIGURE 3 y y=ƒ ϱ Rn ෇ a nϩ1 ϩ a nϩ2 ϩ и и и ഛ y f ͑x͒ dx n an+1 an+2 0 FIGURE 4 n+1 ... Similarly, we see from Figure 4 that x Rn ෇ a nϩ1 ϩ a nϩ2 ϩ и и и ജ y ϱ nϩ1 f ͑x͒ dx 5E-12(pp 756-765) 1/18/06 10:10 AM Page 763 SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 763 So we have proved the following error estimate. 2 Remainder Estimate for the Integral Test Suppose f ͑k͒ ෇ a k, where f is a continuous, positive, decreasing function for x ജ n and ͸ a n is convergent. If Rn ෇ s Ϫ sn , then y ϱ ϱ f ͑x͒ dx ഛ Rn ഛ y f ͑x͒ dx nϩ1 n EXAMPLE 5 (a) Approximate the sum of the series ͸ 1͞n 3 by using the sum of the first 10 terms. Estimate the error involved in this approximation. (b) How many terms are required to ensure that the sum is accurate to within 0.0005? SOLUTION In both parts (a) and (b) we need to know we have y ϱ n ͫ ͬ t 1 1 dx ෇ lim Ϫ 2 tlϱ x3 2x n xnϱ f ͑x͒ dx. With ͩ ෇ lim Ϫ tlϱ 1 1 ϩ 2t 2 2n 2 ͪ f ͑x͒ ෇ 1͞x 3, ෇ 1 2n 2 (a) ϱ ͚ n෇1 1 1 1 1 1 Ϸ s10 ෇ 3 ϩ 3 ϩ 3 ϩ и и и ϩ 3 Ϸ 1.1975 n3 1 2 3 10 According to the remainder estimate in (2), we have R10 ഛ y ϱ 10 1 1 1 dx ෇ ෇ x3 2͑10͒2 200 So the size of the error is at most 0.005. (b) Accuracy to within 0.0005 means that we have to find a value of n such that Rn ഛ 0.0005. Since Rn ഛ y ϱ n 1 1 dx ෇ x3 2n 2 1 Ͻ 0.0005 2n 2 we want Solving this inequality, we get n2 Ͼ 1 ෇ 1000 0.001 or n Ͼ s1000 Ϸ 31.6 We need 32 terms to ensure accuracy to within 0.0005. If we add sn to each side of the inequalities in (2), we get 3 sn ϩ y ϱ nϩ1 ϱ f ͑x͒ dx ഛ s ഛ sn ϩ y f ͑x͒ dx n 5E-12(pp 756-765) 764 ❙❙❙❙ 1/18/06 10:10 AM Page 764 CHAPTER 12 INFINITE SEQUENCES AND SERIES because sn ϩ Rn ෇ s. The inequalities in (3) give a lower bound and an upper bound for s. They provide a more accurate approximation to the sum of the series than the partial sum sn does. ϱ EXAMPLE 6 Use (3) with n ෇ 10 to estimate the sum of the series ͚ n෇1 1 . n3 SOLUTION The inequalities in (3) become s10 ϩ y ϱ 11 1 ϱ 1 3 dx ഛ s ഛ s10 ϩ y 3 dx 10 x x From Example 5 we know that y ϱ n s10 ϩ so 1 1 dx ෇ x3 2n 2 1 1 2 ഛ s ഛ s10 ϩ 2͑11͒ 2͑10͒2 Using s10 Ϸ 1.197532, we get 1.201664 ഛ s ഛ 1.202532 If we approximate s by the midpoint of this interval, then the error is at most half the length of the interval. So ϱ ͚ n෇1 y 1 Ϸ 1.2021 n3 with error Ͻ 0.0005 If we compare Example 6 with Example 5, we see that the improved estimate in (3) can be much better than the estimate s Ϸ sn . To make the error smaller than 0.0005 we had to use 32 terms in Example 5 but only 10 terms in Example 6. y=ƒ Proof of the Integral Test a™ a£ a¢ a∞ 0 1 2 3 4 an 5 ... n x FIGURE 5 y We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series ͸ 1͞n 2 and ͸ 1͞sn. For the general series ͸ a n look at Figures 5 and 6. The area of the first shaded rectangle in Figure 5 is the value of f at the right endpoint of ͓1, 2͔, that is, f ͑2͒ ෇ a 2 . So, comparing the areas of the shaded rectangles with the area under y ෇ f ͑x͒ from 1 to n, we see that y=ƒ 4 an-1 a¡ a™ a£ a¢ 0 1 FIGURE 6 2 3 4 5 ... n a 2 ϩ a 3 ϩ и и и ϩ a n ഛ y f ͑x͒ dx 1 (Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6 shows that n x 5 y n 1 f ͑x͒ dx ഛ a 1 ϩ a 2 ϩ и и и ϩ a nϪ1 5E-12(pp 756-765) 1/18/06 10:10 AM Page 765 SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 765 ϱ (i) If y f ͑x͒ dx is convergent, then (4) gives 1 n ͚a ϱ n ഛ y f ͑x͒ dx ഛ y f ͑x͒ dx i 1 i෇2 1 since f ͑x͒ ജ 0. Therefore n ͚a sn ෇ a 1 ϩ ϱ i ഛ a 1 ϩ y f ͑x͒ dx ෇ M, say 1 i෇2 Since sn ഛ M for all n, the sequence ͕sn ͖ is bounded above. Also snϩ1 ෇ sn ϩ a nϩ1 ജ sn since a nϩ1 ෇ f ͑n ϩ 1͒ ജ 0. Thus, ͕sn ͖ is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (12.1.11). This means that ͸ a n is convergent. (ii) If x1ϱ f ͑x͒ dx is divergent, then x1n f ͑x͒ dx l ϱ as n l ϱ because f ͑x͒ ജ 0. But (5) gives y n 1 nϪ1 ͚a f ͑x͒ dx ഛ i ෇ snϪ1 i෇1 and so snϪ1 l ϱ. This implies that sn l ϱ and so ͸ a n diverges. |||| 12.3 Exercises 11. 1 ϩ ϱ 1 ϱ 1 1.3 Ͻ y 1.3 dx 1 x n ͚ n෇2 1 1 1 1 ϩ ϩ ϩ ϩ иии 8 27 64 125 12. 1 ϩ 1. Draw a picture to show that 1 1 1 1 ϩ ϩ ϩ ϩ иии 2 s2 3 s3 4 s4 5 s5 What can you conclude about the series? ϱ 6 1 5 ͚a f ͑x͒ dx i෇1 ͚ 1 n ϩ4 16. n 2 n ϩ1 18. ϱ 15. n෇1 6 ͚a i 14. ͚ n෇1 x ജ 1 and an ෇ f ͑n͒. By drawing a picture, rank the following three quantities in increasing order: y i 17. i෇2 ͚ n෇1 3–8 Use the Integral Test to determine whether the series is convergent or divergent. ͚ n෇1 ϱ 6. ϱ 1 n4 4. n෇1 Ϫn 7. n෇1 |||| ■ ͚ n෇1 ■ ■ n෇1 ϱ ͚ ne Ϫn ■ ͚ 8. ͚ n෇1 ■ ■ ■ 21. nϩ2 nϩ1 23. ■ 2 n 0.85 ϱ 10. ͚ ͑n n෇1 Ϫ1.4 ϩ 3n Ϫ1.2 ͒ ϱ n෇2 ϱ ͚ n෇1 ■ ■ ϱ 20. 25–28 ϱ 25. ͚ n෇2 ϱ 22. 1 n ϩn 24. ■ ͚ n෇1 ϱ 3 |||| ͚ n෇1 1 n ln n ■ ͚ n෇1 Ϫn 2 ͚ ͚ n෇1 ͚ ne ϱ ■ ϱ n෇1 1 3n ϩ 1 Determine whether the series is convergent or divergent. ϱ 9. 5. n෇1 ■ 9–24 ϱ 1 sn 4 ϱ ͚e ■ ͚ 19. ͚ n෇3 2 ϱ |||| ϱ 3. ϱ 5 Ϫ 2 sn n3 ϱ 13. 2. Suppose f is a continuous positive decreasing function for ͚ n෇3 ■ ■ ■ ■ ■ 5 nϪ2 3n ϩ 2 n͑n ϩ 1͒ 1 n Ϫ 4n ϩ 5 2 ln n n2 n n4 ϩ 1 1 n ln n ln͑ln n͒ ■ ■ ■ Find the values of p for which the series is convergent. 1 n͑ln n͒ p ϱ 26. ͚ n෇3 1 n ln n ͓ln͑ln n͔͒ p ■ 5E-12(pp 766-775) 766 ❙❙❙❙ 1/18/06 Page 766 CHAPTER 12 INFINITE SEQUENCES AND SERIES ϱ 27. 10:54 AM ͚ n͑1 ϩ n ϱ ͒ 2 p 28. n෇1 ■ ■ ■ ͚ n෇1 ■ ■ ■ ■ ln n np ■ CAS ■ ■ ■ ■ 29. The Riemann zeta-function ␨ is defined by ␨ ͑x͒ ෇ ϱ ͚ n෇1 1 nx and is used in number theory to study the distribution of prime numbers. What is the domain of ␨ ? 30. (a) Find the partial sum s10 of the series ͸ϱ 1͞n 4. Estimate the n෇1 error in using s10 as an approximation to the sum of the series. (b) Use (3) with n ෇ 10 to give an improved estimate of the sum. (c) Find a value of n so that sn is within 0.00001 of the sum. 36. (a) Show that the series ͸ϱ ͑ln n͒2͞n 2 is convergent. n෇1 (b) Find an upper bound for the error in the approximation s Ϸ sn . (c) What is the smallest value of n such that this upper bound is less than 0.05? (d) Find sn for this value of n. 37. (a) Use (4) to show that if sn is the nth partial sum of the har- monic series, then sn ഛ 1 ϩ ln n (b) The harmonic series diverges, but very slowly. Use part (a) to show that the sum of the first million terms is less than 15 and the sum of the first billion terms is less than 22. 38. Use the following steps to show that the sequence tn ෇ 1 ϩ 31. (a) Use the sum of the first 10 terms to estimate the sum of the series ͸ϱ 1͞n 2. How good is this estimate? n෇1 (b) Improve this estimate using (3) with n ෇ 10. (c) Find a value of n that will ensure that the error in the approximation s Ϸ sn is less than 0.001. 32. Find the sum of the series ͸ϱ 1͞n 5 correct to three decimal n෇1 has a limit. (The value of the limit is denoted by ␥ and is called Euler’s constant.) (a) Draw a picture like Figure 6 with f ͑x͒ ෇ 1͞x and interpret tn as an area [or use (5)] to show that tn Ͼ 0 for all n. (b) Interpret tn Ϫ tnϩ1 ෇ ͓ln͑n ϩ 1͒ Ϫ ln n͔ Ϫ places. 33. Estimate ͸ϱ nϪ3͞2 to within 0.01. n෇1 34. How many terms of the series ͸ϱ 1͓͞n͑ln n͒ 2͔ would you need n෇2 to add to find its sum to within 0.01? 35. Show that if we want to approximate the sum of the series ͸ϱ nϪ1.001 so that the error is less than 5 in the ninth decimal n෇1 place, then we need to add more than 10 11,301 terms! |||| 12.4 1 1 1 ϩ ϩ и и и ϩ Ϫ ln n 2 3 n 1 nϩ1 as a difference of areas to show that tn Ϫ tnϩ1 Ͼ 0. Therefore, ͕tn ͖ is a decreasing sequence. (c) Use the Monotonic Sequence Theorem to show that ͕tn ͖ is convergent. 39. Find all positive values of b for which the series ͸ϱ b ln n n෇1 converges. The Comparison Tests In the comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent. For instance, the series ϱ 1 ͚ n෇1 1 2 ϩ1 n reminds us of the series ͸ϱ 1͞2 n, which is a geometric series with a ෇ 1 and r ෇ 1 and n෇1 2 2 is therefore convergent. Because the series (1) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality 1 1 Ͻ n 2n ϩ 1 2 shows that our given series (1) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It 5E-12(pp 766-775) 1/18/06 10:54 AM Page 767 SECTION 12.4 THE COMPARISON TESTS ❙❙❙❙ 767 also follows that the sum of the series is less than the sum of the geometric series: ϱ 1 Ͻ1 2 ϩ1 ͚ n n෇1 Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent. ͸ a n and ͸ bn are series with positive terms. ͸ bn is convergent and a n ഛ bn for all n, then ͸ a n is also convergent. If If ͸ bn is divergent and a n ജ bn for all n, then ͸ a n is also divergent. The Comparison Test Suppose that (i) (ii) |||| It is important to keep in mind the distinction between a sequence and a series. A sequence is a list of numbers, whereas a series is a sum. With every series ͸ a n there are associated two sequences: the sequence ͕a n ͖ of terms and the sequence ͕sn ͖ of partial sums. Standard Series for Use with the Comparison Test Proof (i) Let n sn ෇ ͚a n tn ෇ i i෇1 ͚b t෇ i i෇1 ϱ ͚b n n෇1 Since both series have positive terms, the sequences ͕sn ͖ and ͕tn ͖ are increasing ͑snϩ1 ෇ sn ϩ a nϩ1 ജ sn ͒. Also tn l t, so tn ഛ t for all n. Since a i ഛ bi , we have sn ഛ tn . Thus, sn ഛ t for all n. This means that ͕sn ͖ is increasing and bounded above and therefore converges by the Monotonic Sequence Theorem. Thus, ͸ a n converges. (ii) If ͸ bn is divergent, then tn l ϱ (since ͕tn ͖ is increasing). But a i ജ bi so sn ജ tn . Thus, sn l ϱ. Therefore, ͸ a n diverges. In using the Comparison Test we must, of course, have some known series ͸ bn for the purpose of comparison. Most of the time we use either a p-series [͸ 1͞n p converges if p Ͼ 1 and diverges if p ഛ 1; see (12.3.1)] or a geometric series [͸ ar nϪ1 converges if r Ͻ 1 and diverges if r ജ 1; see (12.2.4)]. Խ Խ Խ Խ ϱ EXAMPLE 1 Determine whether the series ͚ n෇1 5 converges or diverges. 2n 2 ϩ 4n ϩ 3 SOLUTION For large n the dominant term in the denominator is 2n 2 so we compare the given series with the series ͸ 5͑͞2n 2 ͒. Observe that 5 5 Ͻ 2n 2 ϩ 4n ϩ 3 2n 2 because the left side has a bigger denominator. (In the notation of the Comparison Test, a n is the left side and bn is the right side.) We know that ϱ ͚ n෇1 5 5 2 ෇ 2n 2 ϱ ͚ n෇1 1 n2 is convergent because it’s a constant times a p-series with p ෇ 2 Ͼ 1. Therefore ϱ ͚ n෇1 5 2n 2 ϩ 4n ϩ 3 is convergent by part (i) of the Comparison Test. 5E-12(pp 766-775) 768 ❙❙❙❙ 1/18/06 10:54 AM Page 768 CHAPTER 12 INFINITE SEQUENCES AND SERIES NOTE 1 Although the condition a n ഛ bn or a n ജ bn in the Comparison Test is given for all n, we need verify only that it holds for n ജ N, where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example. ■ ϱ EXAMPLE 2 Test the series ͚ n෇1 ln n for convergence or divergence. n SOLUTION This series was tested (using the Integral Test) in Example 4 in Section 12.3, but it is also possible to test it by comparing it with the harmonic series. Observe that ln n Ͼ 1 for n ജ 3 and so ln n 1 Ͼ n n nജ3 We know that ͸ 1͞n is divergent ( p-series with p ෇ 1). Thus, the given series is divergent by the Comparison Test. NOTE 2 The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series ■ ϱ ͚ n෇1 1 2 Ϫ1 n The inequality 1 1 Ͼ n 2 Ϫ1 2 n 1 is useless as far as the Comparison Test is concerned because ͸ bn ෇ ͸ ( 2 ) is convergent n and a n Ͼ bn. Nonetheless, we have the feeling that ͸ 1͑͞2 Ϫ 1͒ ought to be convergent 1 n because it is very similar to the convergent geometric series ͸ ( 2 ) . In such cases the following test can be used. n The Limit Comparison Test Suppose that |||| Exercises 40 and 41 deal with the cases c ෇ 0 and c ෇ ϱ. ͸ a n and ͸ bn are series with positive terms. If lim nlϱ an ෇c bn where c is a finite number and c Ͼ 0, then either both series converge or both diverge. Proof Let m and M be positive numbers such that m Ͻ c Ͻ M . Because a n ͞bn is close to c for large n, there is an integer N such that mϽ and so an ϽM bn mbn Ͻ a n Ͻ Mbn when n Ͼ N when n Ͼ N 5E-12(pp 766-775) 1/18/06 10:54 AM Page 769 SECTION 12.4 THE COMPARISON TESTS ❙❙❙❙ 769 If ͸ bn converges, so does ͸ Mbn . Thus, ͸ a n converges by part (i) of the Comparison Test. If ͸ bn diverges, so does ͸ mbn and part (ii) of the Comparison Test shows that ͸ a n diverges. ϱ EXAMPLE 3 Test the series ͚ n෇1 1 for convergence or divergence. 2n Ϫ 1 SOLUTION We use the Limit Comparison Test with an ෇ 1 2n Ϫ 1 bn ෇ 1 2n and obtain lim nlϱ an 1͑͞2 n Ϫ 1͒ 2n 1 ෇ lim ෇ lim n ෇ lim ෇1Ͼ0 n nlϱ nlϱ 2 Ϫ 1 n l ϱ 1 Ϫ 1͞2 n bn 1͞2 Since this limit exists and ͸ 1͞2 n is a convergent geometric series, the given series converges by the Limit Comparison Test. ϱ ͚ EXAMPLE 4 Determine whether the series n෇1 2n 2 ϩ 3n converges or diverges. s5 ϩ n 5 SOLUTION The dominant part of the numerator is 2n 2 and the dominant part of the denom- inator is sn5 ෇ n 5͞2. This suggests taking an ෇ lim nlϱ 2n 2 ϩ 3n s5 ϩ n 5 bn ෇ 2n 2 2 5͞2 ෇ n n 1͞2 an 2n 2 ϩ 3n n 1͞2 2n 5͞2 ϩ 3n 3͞2 ෇ lim ؒ ෇ lim n l ϱ s5 ϩ n 5 nlϱ bn 2 2s5 ϩ n 5 2ϩ ͱ ෇ lim nlϱ 2 3 n 5 ϩ1 n5 ෇ 2ϩ0 ෇1 2s0 ϩ 1 Since ͸ bn ෇ 2 ͸ 1͞n 1͞2 is divergent ( p-series with p ෇ 1 Ͻ 1), the given series diverges 2 by the Limit Comparison Test. Notice that in testing many series we find a suitable comparison series ͸ bn by keeping only the highest powers in the numerator and denominator. Estimating Sums If we have used the Comparison Test to show that a series ͸ a n converges by comparison with a series ͸ bn, then we may be able to estimate the sum ͸ a n by comparing remainders. As in Section 12.3, we consider the remainder Rn ෇ s Ϫ sn ෇ a nϩ1 ϩ a nϩ2 ϩ и и и For the comparison series ͸ bn we consider the corresponding remainder Tn ෇ t Ϫ tn ෇ bnϩ1 ϩ bnϩ2 ϩ и и и 5E-12(pp 766-775) ❙❙❙❙ 770 1/18/06 10:54 AM Page 770 CHAPTER 12 INFINITE SEQUENCES AND SERIES Since a n ഛ bn for all n, we have Rn ഛ Tn . If ͸ bn is a p-series, we can estimate its remainder Tn as in Section 12.3. If ͸ bn is a geometric series, then Tn is the sum of a geometric series and we can sum it exactly (see Exercises 35 and 36). In either case we know that Rn is smaller than Tn . EXAMPLE 5 Use the sum of the first 100 terms to approximate the sum of the series ͸ 1͑͞n 3 ϩ 1͒. Estimate the error involved in this approximation. SOLUTION Since 1 1 Ͻ 3 n ϩ1 n 3 the given series is convergent by the Comparison Test. The remainder Tn for the comparison series ͸ 1͞n 3 was estimated in Example 5 in Section 12.3 using the Remainder Estimate for the Integral Test. There we found that Tn ഛ y ϱ n 1 1 3 dx ෇ x 2n 2 Therefore, the remainder Rn for the given series satisfies Rn ഛ Tn ഛ 1 2n 2 With n ෇ 100 we have R100 ഛ 1 ෇ 0.00005 2͑100͒2 Using a programmable calculator or a computer, we find that ϱ ͚ n෇1 100 1 1 Ϸ ͚ 3 Ϸ 0.6864538 n ϩ1 n෇1 n ϩ 1 3 with error less than 0.00005. |||| 12.4 Exercises 1. Suppose ͸ a n and ͸ bn are series with positive terms and ͸ bn is known to be convergent. (a) If a n Ͼ bn for all n, what can you say about ͸ a n? Why? (b) If a n Ͻ bn for all n, what can you say about ͸ a n? Why? ϱ 7. n෇1 ϱ 3–32 |||| ϱ 3. ͚ n෇1 ϱ 5. ͚ n෇1 ϱ 4. 11. 5 2 ϩ 3n 6. 13. 12. nϪ1 n4n 14. 2 ϩ ͑Ϫ1͒ n n sn 16. 1 sn ϩ 1 18. ͚ n෇1 ϱ ͚ ϱ 1 n Ϫ sn 17. n෇2 n ϩ1 n3 Ϫ 1 ϱ 15. ͚ ϱ 10. ͚ ͚ n෇1 ϱ ͚ n෇1 ͚ n෇1 ϱ 2 2 ͚ n෇1 2 cos n n2 ϩ 1 n෇2 2 n3 ϩ 4 n෇1 ϱ 8. ͚ n෇1 Determine whether the series converges or diverges. 1 n2 ϩ n ϩ 1 nϩ1 n2 ϱ 9. 2. Suppose ͸ a n and ͸ bn are series with positive terms and ͸ bn is known to be divergent. (a) If a n Ͼ bn for all n, what can you say about ͸ a n ? Why? (b) If a n Ͻ bn for all n, what can you say about ͸ a n ? Why? ͚ ͚ n෇0 ϱ ͚ n෇2 ϱ ͚ n෇1 ϱ ͚ n෇1 4 ϩ 3n 2n n2 Ϫ 1 3n 4 ϩ 1 1 ϩ sin n 10 n sn nϪ1 1 sn 3 ϩ 1 1 2n ϩ 3 5E-12(pp 766-775) 1/18/06 10:54 AM Page 771 SECTION 12.5 ALTERNATING SERIES ϱ 19. ͚ n෇1 ϱ 21. ͚ n෇1 ϱ 23. ͚ n෇1 ϱ 25. ͚ n෇1 ϱ 27. ͚ n෇1 ϱ 29. ͚ n෇1 ϱ 31. 20. 1 1 ϩ sn 22. ■ ■ ϱ ϱ 1 ϩ n ϩ n2 s1 ϩ n 2 ϩ n 6 26. 1 n! ͚ n෇1 ͩ ͪ 1 n ͚ n෇3 24. 1ϩ ͚ n෇1 5 ϩ 2n ͑1 ϩ n 2 ͒ 2 ϱ ͚ n෇1 ϱ 2 e Ϫn 28. ͚ n෇1 ϱ 30. ϱ 1 n ■ ͚ n෇1 ͩͪ ͚ sin n෇1 ϱ 2n 1 ϩ 3n 32. ͚ n෇1 ■ ■ ■ ■ ■ 2 39. Prove that if a n ജ 0 and ͸ a n converges, then ͸ a n also converges. nϩ2 ͑n ϩ 1͒ 3 40. (a) Suppose that ͸ an and ͸ bn are series with positive terms and ͸ bn is convergent. Prove that if n 2 Ϫ 5n 3 n ϩnϩ1 lim nlϱ 2n 2 ϩ 7n 3 ͑n 2 ϩ 5n Ϫ 1͒ n 41. (a) Suppose that ͸ an and ͸ bn are series with positive terms and ͸ bn is divergent. Prove that if n! nn lim nlϱ 1 n 1ϩ1͞n ■ 33. ͚ n෇1 ϱ 35. ͚ n෇1 ■ ■ ϱ 34. n෇1 ϱ 1 1 ϩ 2n ■ ͚ 36. ͚ n෇1 ■ ■ ■ ■ ■ an ෇0 bn then ͸ an is also convergent. (b) Use part (a) to show that the series converges. ϱ ϱ ln n ln n (ii) ͚ (i) ͚ 3 n sne n n෇1 n෇1 nϩ5 3 sn 7 ϩ n 2 ■ ■ 42. Give an example of a pair of series ͸ a n and ͸ bn with positive terms where lim n l ϱ ͑a n͞bn͒ ෇ 0 and ͸ bn diverges, but ͸ a n converges. [Compare with Exercise 40.] 1 ϩ cos n n5 43. Show that if a n Ͼ 0 and lim n l ϱ na n n ͑n ϩ 1͒3 n ■ an ෇ϱ bn then ͸ an is also divergent. (b) Use part (a) to show that the series diverges. ϱ ϱ 1 ln n (ii) ͚ (i) ͚ n n෇2 ln n n෇1 ■ |||| Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error. 1 4 n ϩ n2 771 38. For what values of p does the series ͸ϱ 1͑͞n p ln n͒ converge? n෇2 1 ϩ 2n 1 ϩ 3n 33–36 ϱ ❙❙❙❙ 0, then ͸ a n is divergent. ■ ■ ■ 44. Show that if a n Ͼ 0 and ͸ a n is convergent, then ͸ ln͑1 ϩ a n ͒ is convergent. 37. The meaning of the decimal representation of a number 45. If ͸ a n is a convergent series with positive terms, is it true that 0.d1 d2 d3 . . . (where the digit d i is one of the numbers 0, 1, 2, . . . , 9) is that d2 d1 d3 d4 0.d1 d2 d3 d4 . . . ෇ ϩ ϩ иии 2 ϩ 3 ϩ 10 10 10 10 4 46. ͸ sin͑a n ͒ is also convergent? If ͸ a n and ͸ bn are both convergent series with positive terms, is it true that ͸ a n bn is also convergent? Show that this series always converges. |||| 12.5 Alternating Series The convergence tests that we have looked at so far apply only to series with positive terms. In this section and the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance are alternating series, whose terms alternate in sign. An alternating series is a series whose terms are alternately positive and negative. Here are two examples: 1Ϫ Ϫ ϱ 1 ͑Ϫ1͒nϪ1 1 1 1 1 ϩ Ϫ ϩ Ϫ ϩ иии ෇ ͚ 2 3 4 5 6 n n෇1 ϱ 1 2 3 4 5 6 n ϩ Ϫ ϩ Ϫ ϩ Ϫ и и и ෇ ͚ ͑Ϫ1͒n 2 3 4 5 6 7 nϩ1 n෇1 5E-12(pp 766-775) 772 ❙❙❙❙ 1/18/06 10:54 AM Page 772 CHAPTER 12 INFINITE SEQUENCES AND SERIES We see from these examples that the n th term of an alternating series is of the form a n ෇ ͑Ϫ1͒nϪ1bn a n ෇ ͑Ϫ1͒nbn or Խ Խ where bn is a positive number. (In fact, bn ෇ a n .) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. The Alternating Series Test If the alternating series ϱ ͚ ͑Ϫ1͒ nϪ1 bn ෇ b1 Ϫ b2 ϩ b3 Ϫ b4 ϩ b5 Ϫ b6 ϩ и и и ͑bn Ͼ 0͒ n෇1 satisfies (i) bnϩ1 ഛ bn (ii) lim bn ෇ 0 for all n nlϱ then the series is convergent. Before giving the proof let’s look at Figure 1, which gives a picture of the idea behind the proof. We first plot s1 ෇ b1 on a number line. To find s2 we subtract b2 , so s2 is to the left of s1 . Then to find s3 we add b3 , so s3 is to the right of s2 . But, since b3 Ͻ b2 , s3 is to the left of s1 . Continuing in this manner, we see that the partial sums oscillate back and forth. Since bn l 0, the successive steps are becoming smaller and smaller. The even partial sums s2 , s4 , s6 , . . . are increasing and the odd partial sums s1 , s3 , s5 , . . . are decreasing. Thus, it seems plausible that both are converging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separately. b¡ -b™ +b£ -b¢ +b∞ -bß FIGURE 1 0 s™ s¢ sß s s∞ s£ s¡ Proof of the Alternating Series Test We first consider the even partial sums: s2 ෇ b1 Ϫ b2 ജ 0 s4 ෇ s2 ϩ ͑b3 Ϫ b4 ͒ ജ s2 In general since b2 ഛ b1 since b4 ഛ b3 s2n ෇ s2nϪ2 ϩ ͑b2nϪ1 Ϫ b2n ͒ ജ s2nϪ2 since b2n ഛ b2nϪ1 0 ഛ s2 ഛ s4 ഛ s6 ഛ и и и ഛ s2n ഛ и и и Thus But we can also write s2n ෇ b1 Ϫ ͑b2 Ϫ b3 ͒ Ϫ ͑b4 Ϫ b5 ͒ Ϫ и и и Ϫ ͑b2nϪ2 Ϫ b2nϪ1 ͒ Ϫ b2n 5E-12(pp 766-775) 1/18/06 10:55 AM Page 773 SECTION 12.5 ALTERNATING SERIES ❙❙❙❙ 773 Every term in brackets is positive, so s2n ഛ b1 for all n. Therefore, the sequence ͕s2n ͖ of even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Let’s call its limit s, that is, lim s2n ෇ s nlϱ Now we compute the limit of the odd partial sums: lim s2nϩ1 ෇ lim ͑s2n ϩ b2nϩ1 ͒ nlϱ nlϱ ෇ lim s2n ϩ lim b2nϩ1 nlϱ nlϱ ෇sϩ0 [by condition (ii)] ෇s Since both the even and odd partial sums converge to s, we have lim n l ϱ sn ෇ s (see Exercise 72 in Section 12.1) and so the series is convergent. |||| Figure 2 illustrates Example 1 by showing the graphs of the terms a n ෇ ͑Ϫ1͒ nϪ1͞n and the partial sums sn . Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7. In fact, the exact sum of the series is ln 2 Ϸ 0.693 (see Exercise 36). EXAMPLE 1 The alternating harmonic series 1Ϫ ϱ 1 1 1 ͑Ϫ1͒nϪ1 ϩ Ϫ ϩ иии ෇ ͚ 2 3 4 n n෇1 satisfies (i) bnϩ1 Ͻ bn 1 ͕sn ͖ 1 ෇0 n so the series is convergent by the Alternating Series Test. ϱ ͕an ͖ EXAMPLE 2 The series ͚ n෇1 n nlϱ ͑Ϫ1͒n 3n is alternating but 4n Ϫ 1 lim bn ෇ lim nlϱ FIGURE 2 1 1 Ͻ nϩ1 n (ii) lim bn ෇ lim nlϱ 0 because nlϱ 3n ෇ lim nlϱ 4n Ϫ 1 3 4Ϫ 1 n ෇ 3 4 so condition (ii) is not satisfied. Instead, we look at the limit of the nth term of the series: lim a n ෇ lim nlϱ nlϱ ͑Ϫ1͒n 3n 4n Ϫ 1 This limit does not exist, so the series diverges by the Test for Divergence. ϱ EXAMPLE 3 Test the series ͚ ͑Ϫ1͒ nϩ1 n෇1 n2 for convergence or divergence. n3 ϩ 1 SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test. Unlike the situation in Example 1, it is not obvious that the sequence given by bn ෇ n 2͑͞n 3 ϩ 1͒ is decreasing. However, if we consider the related function 5E-12(pp 766-775) 774 ❙❙❙❙ 1/18/06 10:55 AM Page 774 CHAPTER 12 INFINITE SEQUENCES AND SERIES f ͑x͒ ෇ x 2͑͞x 3 ϩ 1͒, we find that x͑2 Ϫ x 3 ͒ ͑x 3 ϩ 1͒2 f Ј͑x͒ ෇ |||| Instead of verifying condition (i) of the Alternating Series Test by computing a derivative, we could verify that bnϩ1 Ͻ bn directly by using the technique of Solution 1 of Example 11 in Section 12.1. Since we are considering only positive x, we see that f Ј͑x͒ Ͻ 0 if 2 Ϫ x 3 Ͻ 0, that is, 3 3 x Ͼ s2. Thus, f is decreasing on the interval (s2, ϱ). This means that f ͑n ϩ 1͒ Ͻ f ͑n͒ and therefore bnϩ1 Ͻ bn when n ജ 2. (The inequality b2 Ͻ b1 can be verified directly but all that really matters is that the sequence ͕bn ͖ is eventually decreasing.) Condition (ii) is readily verified: 1 n 2 lim bn ෇ lim nlϱ nlϱ n ෇ lim nlϱ n3 ϩ 1 1ϩ 1 n3 ෇0 Thus, the given series is convergent by the Alternating Series Test. Estimating Sums A partial sum sn of any convergent series can be used as an approximation to the total sum s, but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using s Ϸ sn is the remainder Rn ෇ s Ϫ sn . The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bnϩ1 , which is the absolute value of the first neglected term. ͸ ͑Ϫ1͒nϪ1bn is the sum of an alternating Alternating Series Estimation Theorem If s ෇ |||| You can see geometrically why the Alternating Series Estimation Theorem is true by looking at Figure 1 (on page 772). Notice that s Ϫ s4 Ͻ b5, s Ϫ s5 Ͻ b6, and so on. Notice also that s lies between any two consecutive partial sums. Խ series that satisfies (i) 0 ഛ bnϩ1 ഛ bn Խ (ii) lim bn ෇ 0 and nlϱ ԽR Խ ෇ Խs Ϫ s Խ ഛ b then n n nϩ1 Proof We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums sn and snϩ1 . It follows that Խs Ϫ s Խ ഛ Խs n nϩ1 ϱ EXAMPLE 4 Find the sum of the series ͚ n෇0 (By definition, 0! ෇ 1.) Խ Ϫ sn ෇ bnϩ1 ͑Ϫ1͒n correct to three decimal places. n! SOLUTION We first observe that the series is convergent by the Alternating Series Test because (i) 1 1 1 ෇ Ͻ ͑n ϩ 1͒! n!͑n ϩ 1͒ n! (ii) 0Ͻ 1 1 1 l 0 as n l ϱ Ͻ l 0 so n! n n! 5E-12(pp 766-775) 1/18/06 10:55 AM Page 775 SECTION 12.5 ALTERNATING SERIES ❙❙❙❙ 775 To get a feel for how many terms we need to use in our approximation, let’s write out the first few terms of the series: s෇ 1 1 1 1 1 1 1 1 Ϫ ϩ Ϫ ϩ Ϫ ϩ Ϫ ϩ иии 0! 1! 2! 3! 4! 5! 6! 7! 1 1 1 1 ෇ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 24 Ϫ 120 ϩ 720 Ϫ 5040 ϩ и и и 2 6 1 1 b 7 ෇ 5040 Ͻ 5000 ෇ 0.0002 Notice that 1 1 1 s6 ෇ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 24 Ϫ 120 ϩ 720 Ϸ 0.368056 2 6 and By the Alternating Series Estimation Theorem we know that Խs Ϫ s Խ ഛ b 6 7 Ͻ 0.0002 This error of less than 0.0002 does not affect the third decimal place, so we have |||| In Section 12.10 we will prove that e x ෇ ͸ϱ x n͞n! for all x, so what we have n෇0 obtained in Example 4 is actually an approximation to the number e Ϫ1. s Ϸ 0.368 correct to three decimal places. | |||| 12.5 NOTE The rule that the error (in using sn to approximate s) is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. The rule does not apply to other types of series. ■ Exercises 1. (a) What is an alternating series? ϱ (b) Under what conditions does an alternating series converge? (c) If these conditions are satisfied, what can you say about the remainder after n terms? 2–20 |||| Test the series for convergence or divergence. 2. Ϫ ϩ Ϫ ϩ Ϫ ϩ и и и 1 3 3. 4 7 2 4 3 5 4 6 5 7 4 4 Ϫ 4 ϩ 4 Ϫ 10 ϩ 11 Ϫ и и и 8 9 1 1 1 1 1 Ϫ ϩ Ϫ ϩ Ϫ иии ln 2 ln 3 ln 4 ln 5 ln 6 5. ͚ n෇1 ϱ ͑Ϫ1͒nϪ1 sn 6. 3n Ϫ 1 7. ͚ ͑Ϫ1͒ n 2n ϩ 1 n෇1 ϱ 9. ͚ n෇1 ͚ ͑Ϫ1͒ nϩ1 n෇1 ϱ 13. ͚ ͑Ϫ1͒ n n෇2 ͚ ͑Ϫ1͒ n෇1 n2 n ϩ4 3 n ln n n sn 1 ϩ 2sn ϱ 12. ͚ ͑Ϫ1͒ nϪ1 n෇1 ϱ 14. ͚ ͑Ϫ1͒ nϪ1 n෇1 ͩͪ ͚ ͑Ϫ1͒ n n෇1 ϱ nn 19. ͚ ͑Ϫ1͒n n! n෇1 ■ 20. ͚ n෇1 ■ sin͑n␲͞2͒ n! ϱ 18. ϱ ■ ͚ n෇1 ␲ 17. ͚ ͑Ϫ1͒n sin n n෇1 ϱ ϱ 2n 8. ͚ ͑Ϫ1͒n 4n 2 ϩ 1 n෇1 10. ϱ 16. ■ ■ ■ ■ ͩ ͪ Ϫ n 5 ͩͪ ␲ n cos n ■ ■ ■ ■ |||| Calculate the first 10 partial sums of the series and graph both the sequence of terms and the sequence of partial sums on the same screen. Estimate the error in using the 10th partial sum to approximate the total sum. ͑Ϫ1͒nϪ1 3n Ϫ 1 ϱ ͑Ϫ1͒ 4n 2 ϩ 1 n෇1 cos n␲ n 3͞4 ; 21–22 ϱ nϩ1 ϱ 11. ͚ n෇1 ϱ ͚ ■ 4. ϱ 15. 21. ͚ n෇1 ■ ■ ϱ ͑Ϫ1͒nϪ1 n 3͞2 ■ 22. ͚ n෇1 ■ ■ ■ 23–26 ■ ■ ͑Ϫ1͒nϪ1 n3 ■ ■ ■ ■ |||| How many terms of the series do we need to add in order to find the sum to the indicated accuracy? ϱ e 1͞n n 23. ln n n 24. ͚ n෇1 ϱ ͚ n෇1 ͑Ϫ1͒nϪ1 n2 ( Խ error Խ Ͻ 0.01) ͑Ϫ1͒nϩ1 n4 ( Խ error Խ Ͻ 0.001) 5E-12(pp 776-785) 776 ❙❙❙❙ 25. ͚ ϱ n෇1 1/18/06 10:58 AM Page 776 CHAPTER 12 INFINITE SEQUENCES AND SERIES ͑Ϫ2͒n n! ϱ ( Խ error Խ Ͻ 0.01) ͚ ͑Ϫ1͒ nϪ1 34. n෇2 ■ ϱ 26. ͚ n෇1 ■ ͑Ϫ1͒n n 4n ■ 27–30 ( Խ error Խ Ͻ 0.002) ■ |||| ■ ■ ■ 27. ͚ n෇1 ϱ 29. ͚ n෇1 ■ ■ ■ ■ ■ ■ ■ ■ Approximate the sum of the series correct to four decimal ϱ ϱ 28. ͑Ϫ1͒ nϪ1 n 2 10 n 30. ■ ͚ n෇1 ϱ ͚ n෇1 ■ ■ ■ ■ ■ ͚ ͑Ϫ1͒ n n 8n n෇1 ϱ ■ ■ ■ ■ n෇1 |||| 12.6 ■ ■ ■ ■ ϱ 33. ͚ n෇1 ͑Ϫ1͒nϪ1 ෇ ln 2 n hn Ϫ ln n l ␥ as n l ϱ and therefore h2n Ϫ ln͑2n͒ l ␥ For what values of p is each series convergent? ͑Ϫ1͒nϪ1 np ■ Let hn and sn be the partial sums of the harmonic and alternating harmonic series. (a) Show that s2n ෇ h2n Ϫ hn. (b) From Exercise 38 in Section 12.3 we have ͑Ϫ1͒ n 3 n n! total sum? Explain. |||| ■ 36. Use the following steps to show that ͸ϱ ͑Ϫ1͒nϪ1͞n an overestimate or an underestimate of the n෇1 ͚ ■ bn , where bn ෇ 1͞n if n is odd and bn ෇ 1͞n 2 if n is even, is divergent. Why does the Alternating Series Test not apply? 31. Is the 50th partial sum s50 of the alternating series 32. ■ 35. Show that the series ͸ ͑Ϫ1͒ ͑Ϫ1͒ nϩ1 n5 ■ 32–34 ■ nϪ1 places. ϱ ■ ͑ln n͒ p n as n l ϱ Use these facts together with part (a) to show that s2n l ln 2 as n l ϱ. ͑Ϫ1͒n nϩp Absolute Convergence and the Ratio and Root Tests Given any series ͸ a n , we can consider the corresponding series ϱ ͚ Խa Խ ෇ Խa Խ ϩ Խa Խ ϩ Խa Խ ϩ и и и n 1 2 3 n෇1 whose terms are the absolute values of the terms of the original series. |||| We have convergence tests for series with positive terms and for alternating series. But what if the signs of the terms switch back and forth irregularly? We will see in Example 3 that the idea of absolute convergence sometimes helps in such cases. 1 Definition A series ͸ a n is called absolutely convergent if the series of absolute values ͸ a n is convergent. Խ Խ Notice that if ͸ a n is a series with positive terms, then a n ෇ a n and so absolute convergence is the same as convergence in this case. Խ Խ EXAMPLE 1 The series ϱ ͚ n෇1 ͑Ϫ1͒nϪ1 1 1 1 ෇ 1 Ϫ 2 ϩ 2 Ϫ 2 ϩ иии n2 2 3 4 is absolutely convergent because ϱ ͚ n෇1 Ϳ Ϳ ϱ ͑Ϫ1͒nϪ1 1 1 1 1 ෇ ͚ 2 ෇ 1 ϩ 2 ϩ 2 ϩ 2 ϩ иии 2 n 2 3 4 n෇1 n is a convergent p-series ( p ෇ 2). 5E-12(pp 776-785) 1/18/06 10:58 AM Page 777 SECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 777 EXAMPLE 2 We know that the alternating harmonic series ϱ ͚ n෇1 ͑Ϫ1͒nϪ1 1 1 1 ෇ 1 Ϫ ϩ Ϫ ϩ иии n 2 3 4 is convergent (see Example 1 in Section 12.5), but it is not absolutely convergent because the corresponding series of absolute values is Ϳ ϱ ͚ n෇1 Ϳ ϱ ͑Ϫ1͒nϪ1 1 1 1 1 ෇ ͚ ෇ 1 ϩ ϩ ϩ ϩ иии n 2 3 4 n෇1 n which is the harmonic series ( p-series with p ෇ 1) and is therefore divergent. 2 Definition A series ͸ a n is called conditionally convergent if it is convergent but not absolutely convergent. Example 2 shows that the alternating harmonic series is conditionally convergent. Thus, it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence. ͸ a n is absolutely convergent, then it is convergent. Theorem If a series 3 Proof Observe that the inequality Խ Խ Խ Խ 0 ഛ an ϩ an ഛ 2 an is true because a n is either a n or Ϫa n . If ͸ a n is absolutely convergent, then ͸ a n is convergent, so ͸ 2 a n is convergent. Therefore, by the Comparison Test, ͸ (a n ϩ a n ) is convergent. Then Խ Խ Խ Խ Խ Խ Խ Խ ͚a n Խ Խ) Ϫ ͚ Խ a Խ ෇ ͚ (a n ϩ a n n is the difference of two convergent series and is therefore convergent. |||| Figure 1 shows the graphs of the terms a n and partial sums sn of the series in Example 3. Notice that the series is not alternating but has positive and negative terms. EXAMPLE 3 Determine whether the series ϱ ͚ n෇1 cos n cos 1 cos 2 cos 3 ෇ ϩ ϩ ϩ иии 2 2 2 n 1 2 32 is convergent or divergent. 0.5 ͕sn ͖ ͕ an ͖ 0 FIGURE 1 n SOLUTION This series has both positive and negative terms, but it is not alternating. (The first term is positive, the next three are negative, and the following three are positive. The signs change irregularly.) We can apply the Comparison Test to the series of absolute values ϱ ϱ cos n cos n ͚ n 2 ෇ n෇1 n 2 ͚ n෇1 Ϳ Ϳ Խ Խ Խ Since cos n ഛ 1 for all n, we have Խ cos n Խ ഛ n2 1 n2 Խ 5E-12(pp 776-785) 778 ❙❙❙❙ 1/18/06 10:58 AM Page 778 CHAPTER 12 INFINITE SEQUENCES AND SERIES We know that ͸ 1͞n 2 is convergent ( p-series with p ෇ 2) and therefore ͸ cos n ͞n 2 is convergent by the Comparison Test. Thus, the given series ͸ ͑cos n͒͞n 2 is absolutely convergent and therefore convergent by Theorem 3. Խ Խ The following test is very useful in determining whether a given series is absolutely convergent. The Ratio Test Ϳ Ϳ ϱ a nϩ1 ෇ L Ͻ 1, then the series ͚ a n is absolutely convergent nlϱ an n෇1 (and therefore convergent). (i) If lim Ϳ Ϳ Ϳ Ϳ ϱ a nϩ1 a nϩ1 ෇ L Ͼ 1 or lim ෇ ϱ, then the series ͚ a n nlϱ nlϱ an an n෇1 is divergent. (ii) If lim Ϳ Ϳ a nϩ1 ෇ 1, the Ratio Test is inconclusive; that is, no conclusion can nlϱ an be drawn about the convergence or divergence of ͸ a n. (iii) If lim Proof (i) The idea is to compare the given series with a convergent geometric series. Since L Ͻ 1, we can choose a number r such that L Ͻ r Ͻ 1. Since lim nlϱ Խ Խ Ϳ Ϳ a nϩ1 ෇L an LϽr and the ratio a nϩ1͞a n will eventually be less than r ; that is, there exists an integer N such that a nϩ1 whenever n ജ N Ͻr an Ϳ Ϳ or, equivalently, Խa Խ Ͻ Խa Խr 4 nϩ1 whenever n ജ N n Putting n successively equal to N , N ϩ 1, N ϩ 2, . . . in (4), we obtain Խa Խ Ͻ Խa Խr Խa Խ Ͻ Խa Խr Ͻ Խa Խr Խa Խ Ͻ Խa Խr Ͻ Խa Խr Nϩ1 N Nϩ2 Nϩ1 N Nϩ3 Nϩ2 N 2 3 and, in general, Խa Խ Ͻ Խa Խr 5 Nϩk N k for all k ജ 1 Now the series ϱ ͚ Խa Խr N k෇1 k Խ Խ Խ Խ Խ Խ ෇ aN r ϩ aN r 2 ϩ aN r 3 ϩ и и и 5E-12(pp 776-785) 1/18/06 10:58 AM Page 779 SECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 779 is convergent because it is a geometric series with 0 Ͻ r Ͻ 1. So the inequality (5), together with the Comparison Test, shows that the series ϱ ϱ ͚ Խa Խ ෇ ͚ Խa Խ ෇ Խa Խ ϩ Խa Խ ϩ Խa Խ ϩ и и и n n෇Nϩ1 Nϩk Nϩ1 Nϩ2 Nϩ3 k෇1 is also convergent. It follows that the series ͸ϱ a n is convergent. (Recall that a finite n෇1 number of terms doesn’t affect convergence.) Therefore, ͸ a n is absolutely convergent. (ii) If a nϩ1͞a n l L Ͼ 1 or a nϩ1͞a n l ϱ, then the ratio a nϩ1͞a n will eventually be greater than 1; that is, there exists an integer N such that Խ Խ Խ Խ Խ Խ Խ Ϳ Ϳ a nϩ1 Ͼ1 an Խ Խ whenever n ജ N Խ Խ Խ This means that a nϩ1 Ͼ a n whenever n ജ N and so lim a n 0 nlϱ Therefore, ͸ a n diverges by the Test for Divergence. Խ Խ Part (iii) of the Ratio Test says that if lim n l ϱ a nϩ1͞a n ෇ 1, the test gives no information. For instance, for the convergent series ͸ 1͞n 2 we have NOTE ■ 1 a nϩ1 ͑n ϩ 1͒2 n2 ෇ ෇ ෇ an 1 ͑n ϩ 1͒2 n2 Ϳ Ϳ 1 ͩ ͪ as n l ϱ l1 2 1 1ϩ n whereas for the divergent series ͸ 1͞n we have 1 a nϩ1 nϩ1 n 1 ෇ ෇ ෇ l1 an 1 nϩ1 1 1ϩ n n Ϳ Ϳ as n l ϱ Therefore, if lim n l ϱ a nϩ1͞a n ෇ 1, the series ͸ a n might converge or it might diverge. In this case the Ratio Test fails and we must use some other test. Խ Խ ϱ EXAMPLE 4 Test the series ͚ ͑Ϫ1͒ n n෇1 |||| ESTIMATING SUMS In the last three sections we used various methods for estimating the sum of a series—the method depended on which test was used to prove convergence. What about series for which the Ratio Test works? There are two possibilities: If the series happens to be an alternating series, as in Example 4, then it is best to use the methods of Section 12.5. If the terms are all positive, then use the special methods explained in Exercise 34. n3 for absolute convergence. 3n SOLUTION We use the Ratio Test with a n ෇ ͑Ϫ1͒n n 3͞3 n: Ϳ Ϳ a nϩ1 an ͑Ϫ1͒nϩ1͑n ϩ 1͒3 3 nϩ1 ͑n ϩ 1͒3 3 n ෇ ෇ ؒ 3 n 3 ͑Ϫ1͒ n 3 nϩ1 n n 3 | ෇ 1 3 | ͩ ͪ ͩ ͪ nϩ1 n 3 ෇ 1 3 1ϩ 1 n 3 l 1 Ͻ1 3 Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent. 5E-12(pp 776-785) 780 ❙❙❙❙ 1/18/06 10:58 AM Page 780 CHAPTER 12 INFINITE SEQUENCES AND SERIES ϱ ͚ EXAMPLE 5 Test the convergence of the series n෇1 nn . n! SOLUTION Since the terms a n ෇ n ͞n! are positive, we don’t need the absolute value signs. n a nϩ1 ͑n ϩ 1͒nϩ1 n! ͑n ϩ 1͒͑n ϩ 1͒n n! ෇ ؒ n ෇ ؒ n an ͑n ϩ 1͒! n ͑n ϩ 1͒n! n ෇ ͩ ͪ ͩ ͪ nϩ1 n n n 1 n ෇ 1ϩ as n l ϱ le (See Equation 7.4.9 or 7.4*.9.) Since e Ͼ 1, the given series is divergent by the Ratio Test. NOTE Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence. Since ■ an ෇ nn n ؒ n ؒ n ؒ иии ؒ n ෇ ജn n! 1 ؒ 2 ؒ 3 ؒ иии ؒ n it follows that a n does not approach 0 as n l ϱ. Therefore, the given series is divergent by the Test for Divergence. The following test is convenient to apply when n th powers occur. Its proof is similar to the proof of the Ratio Test and is left as Exercise 38. The Root Test Խ Խ ϱ ͚a n (i) If lim s a n ෇ L Ͻ 1, then the series nlϱ n is absolutely convergent n෇1 (and therefore convergent). Խ Խ Խ Խ n n (ii) If lim s a n ෇ L Ͼ 1 or lim s a n ෇ ϱ, then the series nlϱ nlϱ ϱ ͚a n is divergent. n෇1 Խ Խ n (iii) If lim s a n ෇ 1, the Root Test is inconclusive. nlϱ Խ Խ n If lim n l ϱ s a n ෇ 1, then part (iii) of the Root Test says that the test gives no information. The series ͸ a n could converge or diverge. (If L ෇ 1 in the Ratio Test, don’t try the Root Test because L will again be 1.) ϱ ͚ EXAMPLE 6 Test the convergence of the series n෇1 SOLUTION an ෇ Խ Խ n s an ͩ 2n ϩ 3 3n ϩ 2 ͩ 2n ϩ 3 3n ϩ 2 ͪ n . ͪ n 3 2n ϩ 3 n 2 ෇ ෇ l Ͻ1 3n ϩ 2 3 2 3ϩ n 2ϩ Thus, the given series converges by the Root Test. 5E-12(pp 776-785) 1/18/06 10:59 AM Page 781 SECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 781 Rearrangements The question of whether a given convergent series is absolutely convergent or conditionally convergent has a bearing on the question of whether infinite sums behave like finite sums. If we rearrange the order of the terms in a finite sum, then of course the value of the sum remains unchanged. But this is not always the case for an infinite series. By a rearrangement of an infinite series ͸ a n we mean a series obtained by simply changing the order of the terms. For instance, a rearrangement of ͸ a n could start as follows: a 1 ϩ a 2 ϩ a 5 ϩ a 3 ϩ a 4 ϩ a 15 ϩ a 6 ϩ a 7 ϩ a 20 ϩ и и и It turns out that if ͸ a n is an absolutely convergent series with sum s, then any rearrangement of ͸ a n has the same sum s. However, any conditionally convergent series can be rearranged to give a different sum. To illustrate this fact let’s consider the alternating harmonic series 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ 1 Ϫ 1 ϩ и и и ෇ ln 2 2 3 4 5 6 7 8 6 1 (See Exercise 36 in Section 12.5.) If we multiply this series by 2 , we get 1 2 Ϫ 1 ϩ 1 Ϫ 1 ϩ и и и ෇ 1 ln 2 4 6 8 2 Inserting zeros between the terms of this series, we have |||| Adding these zeros does not affect the sum of the series; each term in the sequence of partial sums is repeated, but the limit is the same. 0 ϩ 1 ϩ 0 Ϫ 1 ϩ 0 ϩ 1 ϩ 0 Ϫ 1 ϩ и и и ෇ 1 ln 2 2 4 6 8 2 7 Now we add the series in Equations 6 and 7 using Theorem 12.2.8: 1 ϩ 1 Ϫ 1 ϩ 1 ϩ 1 Ϫ 1 ϩ и и и ෇ 3 ln 2 3 2 5 7 4 2 8 Notice that the series in (8) contains the same terms as in (6), but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are different. In fact, Riemann proved that if ͸ a n is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of ͸ a n that has a sum equal to r. A proof of this fact is outlined in Exercise 40. |||| 12.6 Exercises 1. What can you say about the series ͸ a n in each of the following cases? (a) lim nlϱ Ϳ Ϳ Ϳ Ϳ a nϩ1 ෇8 an (b) lim nlϱ Ϳ Ϳ a nϩ1 ෇ 0.8 an a nϩ1 ෇1 (c) lim nlϱ an 2–28 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. |||| ϱ 2. ͚ n෇1 n2 2n ϱ ϱ ͑Ϫ10͒ n n! 4. ͑Ϫ1͒ 4 sn 6. n 7. ͚ ͑Ϫ1͒n 5ϩn n෇1 8. 3. ͚ n෇0 ϱ 5. ͚ n෇1 9. ͚ n෇1 ϱ 11. ͚ n෇1 nϪ1 n෇1 ϱ nϩ1 ͚ n෇1 ϱ ϱ ͚ ͑Ϫ1͒ ͑Ϫ1͒n n4 ϱ ͚ ͑Ϫ1͒ nϪ1 n෇1 ϱ 1 ͑2n͒! 10. ͑Ϫ1͒ne 1͞n n3 12. ͚e Ϫn n! n෇1 ϱ ͚ n෇1 2n n4 sin 4n 4n n n2 ϩ 1 5E-12(pp 776-785) 782 ❙❙❙❙ 13. ͚ ϱ n෇1 ϱ 15. ͚ n෇1 ϱ 17. ͚ n෇2 ϱ 19. ͚ n෇1 ͚ n෇1 ϱ 23. ͚ n෇1 ͚ n෇1 10 n ͑n ϩ 1͒4 2nϩ1 16. ϱ ͚ 18. n෇1 ϱ ͚ 20. n෇2 ϱ ͚ 22. 1ϩ3n n2 ϩ 1 2n 2 ϩ 1 n෇2 ͪ ϱ n ͚ 24. n෇1 34. Let ͸ a n be a series with positive terms and let rn ෇ a nϩ1 ͞a n. n 22n n! Suppose that lim n l ϱ rn ෇ L Ͻ 1, so ͸ a n converges by the Ratio Test. As usual, we let Rn be the remainder after n terms, that is, 3 Ϫ cos n n 2͞3 Ϫ 2 Rn ෇ a nϩ1 ϩ a nϩ2 ϩ a nϩ3 ϩ и и и n! nn (a) If ͕rn ͖ is a decreasing sequence and rnϩ1 Ͻ 1, show, by summing a geometric series, that ͑Ϫ1͒ ͑ln n͒n n ͑Ϫ1͒n n ln n (b) If ͕rn ͖ is an increasing sequence, show that ͑Ϫ1͒n ͑arctan n͒n Rn ഛ n ■ cise 34 to estimate the error in using s5 as an approximation to the sum of the series. (b) Find a value of n so that sn is within 0.00005 of the sum. Use this value of n to approximate the sum of the series. 36. Use the sum of the first 10 terms to approximate the sum of the series ϱ ͚ n෇1 2 n n! 5 ؒ 8 ؒ 11 ؒ и и и ؒ ͑3n ϩ 2͒ ■ ■ ■ ■ ■ 37. Prove that if ͸ a n is absolutely convergent, then ■ ■ ■ ■ Ϳ͚ Ϳ ■ ϱ a nϩ1 ෇ 5n ϩ 1 an 4n ϩ 3 Determine whether ͸ a n converges or diverges. 30. A series ͸ a n is defined by the equations a1 ෇ 1 2 ϩ cos n a nϩ1 ෇ an sn Determine whether ͸ a n converges or diverges. 31. For which of the following series is the Ratio Test inconclusive (that is, it fails to give a definite answer)? ϱ ϱ 1 n (a) ͚ 3 (b) ͚ n n෇1 n n෇1 2 ϱ (c) ͚ n෇1 ϱ ͑Ϫ3͒nϪ1 sn ͚ (d) n෇1 sn 1 ϩ n2 32. For which positive integers k is the following series convergent? ϱ ͚ n෇1 n 2n Use Exercise 34 to estimate the error. 29. The terms of a series are defined recursively by the equations a1 ෇ 2 a nϩ1 1ϪL 35. (a) Find the partial sum s5 of the series ͸ϱ 1͞n2 n. Use Exern෇1 2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒ n! ͚ ͑Ϫ1͒ a nϩ1 1 Ϫ rnϩ1 Rn ഛ 1ؒ3ؒ5 1ؒ3ؒ5ؒ7 1ؒ3 ϩ Ϫ ϩ иии 3! 5! 7! 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ ϩ ͑Ϫ1͒nϪ1 ϩ иии ͑2n Ϫ 1͒! n෇1 ■ ͚ n෇1 nn ͩ nϩ1 ϱ cos͑n␲͞3͒ n! 3 ͚ ͑Ϫ1͒ n෇1 ͑Ϫ1͒n ln n ϱ 28. ϱ 14. 2 2ؒ6 2 ؒ 6 ؒ 10 2 ؒ 6 ؒ 10 ؒ 14 ϩ ϩ ϩ ϩ иии 5 5ؒ8 5 ؒ 8 ؒ 11 5 ؒ 8 ؒ 11 ؒ 14 ϱ 27. Page 782 n͑Ϫ3͒n 4 nϪ1 25. 1 Ϫ 26. 10:59 AM CHAPTER 12 INFINITE SEQUENCES AND SERIES ϱ 21. 1/18/06 ͑n!͒ ͑kn͒! 2 33. (a) Show that ͸ϱ x n͞n! converges for all x. n෇0 (b) Deduce that lim n l ϱ x n͞n! ෇ 0 for all x. an ഛ n෇1 ϱ ͚ Խa Խ n n෇1 38. Prove the Root Test. [Hint for part (i): Take any number r such that L Ͻ r Ͻ 1 and use the fact that there is an integer N n such that s a n Ͻ r whenever n ജ N .] Խ Խ ϩ 39. Given any series ͸ a n we define a series ͸ a n whose terms are Ϫ all the positive terms of ͸ a n and a series ͸ a n whose terms are all the negative terms of ͸ a n. To be specific, we let aϩ ෇ n Խ Խ an ϩ an 2 aϪ ෇ n Խ Խ a n Ϫ an 2 Ϫ Notice that if a n Ͼ 0, then aϩ ෇ a n and a n ෇ 0, whereas if n Ϫ ෇ a and aϩ ෇ 0. a n Ͻ 0, then a n n n (a) If ͸ a n is absolutely convergent, show that both of the series ϩ Ϫ ͸ a n and ͸ a n are convergent. (b) If ͸ a n is conditionally convergent, show that both of the ϩ Ϫ series ͸ a n and ͸ a n are divergent. 40. Prove that if ͸ a n is a conditionally convergent series and r is any real number, then there is a rearrangement of ͸ a n whose sum is r . [Hints: Use the notation of Exercise 39. Take just enough positive terms aϩ so that their sum is greater n than r . Then add just enough negative terms aϪ so that the n cumulative sum is less than r . Continue in this manner and use Theorem 12.2.6.] 5E-12(pp 776-785) 1/18/06 10:59 AM Page 783 SECTION 12.7 STRATEGY FOR TESTING SERIES |||| 12.7 ❙❙❙❙ 783 Strategy for Testing Series We now have several ways of testing a series for convergence or divergence; the problem is to decide which test to use on which series. In this respect testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given series, but you may find the following advice of some use. It is not wise to apply a list of the tests in a specific order until one finally works. That would be a waste of time and effort. Instead, as with integration, the main strategy is to classify the series according to its form. 1. If the series is of the form ͸ 1͞n p, it is a p-series, which we know to be conver- gent if p Ͼ 1 and divergent if p ഛ 1. 2. If the series has the form ͸ ar nϪ1 or ͸ ar n, it is a geometric series, which converges if r Ͻ 1 and diverges if r ജ 1. Some preliminary algebraic manipulation may be required to bring the series into this form. Խ Խ Խ Խ 3. If the series has a form that is similar to a p-series or a geometric series, then one of the comparison tests should be considered. In particular, if a n is a rational function or algebraic function of n (involving roots of polynomials), then the series should be compared with a p-series. Notice that most of the series in Exercises 12.4 have this form. (The value of p should be chosen as in Section 12.4 by keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if ͸ a n has some negative terms, then we can apply the Comparison Test to ͸ a n and test for absolute convergence. If you can see at a glance that lim n l ϱ a n 0, then the Test for Divergence should be used. If the series is of the form ͸ ͑Ϫ1͒nϪ1bn or ͸ ͑Ϫ1͒nbn , then the Alternating Series Test is an obvious possibility. Series that involve factorials or other products (including a constant raised to the n th power) are often conveniently tested using the Ratio Test. Bear in mind that a nϩ1͞a n l 1 as n l ϱ for all p-series and therefore all rational or algebraic functions of n. Thus, the Ratio Test should not be used for such series. If a n is of the form ͑bn ͒n, then the Root Test may be useful. If a n ෇ f ͑n͒, where x1ϱ f ͑x͒ dx is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied). Խ Խ 4. 5. 6. Խ 7. 8. Խ In the following examples we don’t work out all the details but simply indicate which tests should be used. ϱ ͚ EXAMPLE 1 n෇1 Since a n l 1 2 ϱ EXAMPLE 2 ͚ n෇1 nϪ1 2n ϩ 1 0 as n l ϱ, we should use the Test for Divergence. sn 3 ϩ 1 3n 3 ϩ 4n 2 ϩ 2 Since a n is an algebraic function of n, we compare the given series with a p-series. 5E-12(pp 776-785) ❙❙❙❙ 784 1/18/06 11:00 AM Page 784 CHAPTER 12 INFINITE SEQUENCES AND SERIES The comparison series for the Limit Comparison Test is ͸ bn , where bn ෇ ϱ n 3͞2 1 sn 3 ෇ ෇ 3 3 3n 3n 3n 3͞2 ͚ ne EXAMPLE 3 Ϫn 2 n෇1 2 Since the integral x1ϱ xeϪx dx is easily evaluated, we use the Integral Test. The Ratio Test also works. ϱ ͚ ͑Ϫ1͒ n EXAMPLE 4 n෇1 n3 n4 ϩ 1 Since the series is alternating, we use the Alternating Series Test. ϱ ͚ EXAMPLE 5 k෇1 2k k! Since the series involves k!, we use the Ratio Test. ϱ ͚ EXAMPLE 6 n෇1 1 2 ϩ 3n Since the series is closely related to the geometric series ͸ 1͞3 n, we use the Comparison Test. |||| 12.7 1–38 |||| ϱ 1. ͚ n෇1 ϱ Exercises n Ϫ1 n2 ϩ n ϱ 2 2. 1 n ϩn 4. ͑Ϫ3͒nϩ1 2 3n 6. 1 nsln n 8. n෇1 5. ͚ n෇1 ϱ 7. ͚ n෇2 ϱ 9. ͚ ϱ ͚ n෇2 ϱ 13. ͚ n෇1 ϱ 15. ͚ n෇0 ϱ ͚ k෇1 ϱ k 2 e Ϫk 10. ͚ ͩ ͑Ϫ1͒ nϩ1 n ln n ϱ nϪ1 n2 ϩ n 3n 1 ϩ 8n ͪ 21. 23. ϱ 27. n! 2 ؒ 5 ؒ 8 ؒ и и и ؒ ͑3n ϩ 2͒ 16. n෇1 ϱ n ͚ n෇1 ϱ 1͞n 18. ͚ n෇2 ϱ 24. ϱ tanϪ1n n sn 30. ͚ 5k 3k ϩ 4k 32. sin͑1͞n͒ sn 34. ͚ ϱ 31. k෇1 33. ͚ n෇1 n2 ϩ 1 n3 ϩ 1 35. ͑Ϫ1͒nϪ1 sn Ϫ 1 37. ϱ ͚ n෇1 ϱ kϩ5 5k sn 2 Ϫ 1 n ϩ 2n 2 ϩ 5 3 cos͑n͞2͒ n 2 ϩ 4n n2 ϩ 1 5n e 1͞n n2 ϱ ͚ ͑Ϫ1͒ j j෇1 ϱ ͚ n෇1 ϱ n ͚ n෇1 n2 ͚ (s2 Ϫ 1) n෇1 ͚ n෇1 ͩ ͪ n ͚ n෇1 n nϩ1 ϱ ͚ n෇1 28. ͚ ͚ n෇1 k ln k ͑k ϩ 1͒3 ϱ 29. ͚ sin n 22. 26. ͚ n෇1 ϱ 14. ϱ ͑Ϫ2͒2n nn ϱ 25. ͚ k෇1 n! 2 en ϱ n n 2 ϩ 25 ϱ 20. n෇1 3 n ln n sn ͚ tan͑1͞n͒ n෇1 ͚ ͑Ϫ1͒ n ϱ k෇1 n 2eϪn n෇1 3nn 2 n! ͚ n෇1 n 2 k k! ͑k ϩ 2͒! ϱ 12. ͚ ͑Ϫ1͒ n෇1 n෇1 ͚ ͑Ϫ1͒ 2 n෇1 ͚ n෇1 ϱ 17. nϪ1 ϱ k෇1 11. ͚ ͑Ϫ1͒ n෇1 19. nϪ1 n2 ϩ n ϱ ͚ 2 ͚ n෇1 ϱ 3. ϱ Test the series for convergence or divergence. ϱ 36. ͚ n෇2 ͑2n͒n n 2n 1 n ϩ n cos2 n 1 ͑ln n͒ln n ϱ 38. sj jϩ5 ͚ (s2 Ϫ 1) n n෇1 5E-12(pp 776-785) 1/18/06 11:00 AM Page 785 SECTION 12.8 POWER SERIES |||| 12.8 ❙❙❙❙ 785 Power Series A power series is a series of the form ϱ ͚cx 1 n n ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и n෇0 where x is a variable and the cn’s are constants called the coefficients of the series. For each fixed x, the series (1) is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function |||| TRIGONOMETRIC SERIES A power series is a series in which each term is a power function. A trigonometric series ϱ ͚ ͑a n cos nx ϩ bn sin nx͒ f ͑x͒ ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ и и и ϩ cn x n ϩ и и и whose domain is the set of all x for which the series converges. Notice that f resembles a polynomial. The only difference is that f has infinitely many terms. For instance, if we take cn ෇ 1 for all n, the power series becomes the geometric series n෇0 is a series whose terms are trigonometric functions. This type of series is discussed on the web site www.stewartcalculus.com Click on Additional Topics and then on Fourier Series. ϱ ͚x n ෇ 1 ϩ x ϩ x2 ϩ и и и ϩ xn ϩ и и и n෇0 Խ Խ which converges when Ϫ1 Ͻ x Ͻ 1 and diverges when x ജ 1 (see Equation 12.2.5). More generally, a series of the form ϱ 2 ͚ c ͑x Ϫ a͒ n n ෇ c0 ϩ c1͑x Ϫ a͒ ϩ c2͑x Ϫ a͒2 ϩ и и и n෇0 is called a power series in ͑x Ϫ a͒ or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to n ෇ 0 in Equations 1 and 2 we have adopted the convention that ͑x Ϫ a͒0 ෇ 1 even when x ෇ a. Notice also that when x ෇ a all of the terms are 0 for n ജ 1 and so the power series (2) always converges when x ෇ a. ϱ ͚ n!x EXAMPLE 1 For what values of x is the series n convergent? n෇0 SOLUTION We use the Ratio Test. If we let a n , as usual, denote the nth term of the series, then a n ෇ n!x n. If x 0, we have lim nlϱ Ϳ Ϳ Ϳ Ϳ anϩ1 ͑n ϩ 1͒!x nϩ1 ෇ lim ෇ lim ͑n ϩ 1͒ x ෇ ϱ nlϱ nlϱ an n!x n By the Ratio Test, the series diverges when x when x ෇ 0. 0. Thus, the given series converges only ϱ EXAMPLE 2 For what values of x does the series Խ Խ ͚ n෇1 SOLUTION Let a n ෇ ͑x Ϫ 3͒ ͞n. Then ͑x Ϫ 3͒n converge? n n Ϳ Ϳ Ϳ n anϩ1 ͑x Ϫ 3͒nϩ1 ෇ ؒ an nϩ1 ͑x Ϫ 3͒n ෇ 1 1ϩ 1 n Խx Ϫ 3Խ Խ Ϳ l xϪ3 Խ as n l ϱ 5E-12(pp 786-795) 786 ❙❙❙❙ 1/18/06 10:36 AM Page 786 CHAPTER 12 INFINITE SEQUENCES AND SERIES By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when x Ϫ 3 Ͻ 1 and divergent when x Ϫ 3 Ͼ 1. Now Խ Խ Խ Խx Ϫ 3Խ Ͻ 1 &? Խ Ϫ1 Ͻ x Ϫ 3 Ͻ 1 2ϽxϽ4 &? so the series converges when 2 Ͻ x Ͻ 4 and diverges when x Ͻ 2 or x Ͼ 4. The Ratio Test gives no information when x Ϫ 3 ෇ 1 so we must consider x ෇ 2 and x ෇ 4 separately. If we put x ෇ 4 in the series, it becomes ͸ 1͞n, the harmonic series, which is divergent. If x ෇ 2, the series is ͸ ͑Ϫ1͒n͞n, which converges by the Alternating Series Test. Thus, the given power series converges for 2 ഛ x Ͻ 4. Խ Խ We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel (1784–1846), and the function given in Exercise 33 is another example of a Bessel function. In fact, these functions first arose when Bessel solved Kepler’s equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the temperature distribution in a circular plate and the shape of a vibrating drumhead (see the photographs on page 736). EXAMPLE 3 Find the domain of the Bessel function of order 0 defined by J0͑x͒ ෇ ϱ ͚ n෇0 ͑Ϫ1͒n x 2n 2 2n͑n!͒2 SOLUTION Let a n ෇ ͑Ϫ1͒n x 2n͓͞2 2n͑n!͒2͔. Then Ϳ Ϳ Ϳ anϩ1 ͑Ϫ1͒nϩ1x 2͑nϩ1͒ 2 2n͑n!͒2 ෇ 2͑nϩ1͒ 2 ؒ an 2 ͓͑n ϩ 1͒!͔ ͑Ϫ1͒nx 2n ෇ ෇ 2 2nϩ2 Ϳ x 2nϩ2 2 2n͑n!͒2 2 2 ؒ ͑n ϩ 1͒ ͑n!͒ x 2n x2 l 0Ͻ1 4͑n ϩ 1͒2 for all x Thus, by the Ratio Test, the given series converges for all values of x. In other words, the domain of the Bessel function J0 is ͑Ϫϱ, ϱ͒ ෇ ‫.ޒ‬ Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we define the Bessel function in Example 3 as the sum of a series we mean that, for every real number x, n J0͑x͒ ෇ lim sn͑x͒ nlϱ where sn͑x͒ ෇ ͚ i෇0 ͑Ϫ1͒ix 2i 2 2i͑i!͒2 The first few partial sums are s0͑x͒ ෇ 1 s3͑x͒ ෇ 1 Ϫ s1͑x͒ ෇ 1 Ϫ x2 x4 x6 ϩ Ϫ 4 64 2304 x2 4 s2͑x͒ ෇ 1 Ϫ s4͑x͒ ෇ 1 Ϫ x2 x4 ϩ 4 64 x2 x4 x6 x8 ϩ Ϫ ϩ 4 64 2304 147,456 5E-12(pp 786-795) 1/18/06 10:36 AM Page 787 SECTION 12.8 POWER SERIES y s™ s¸ 1 s¢ 0 x 1 s¡ s£ ❙❙❙❙ 787 Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function J0 , but notice that the approximations become better when more terms are included. Figure 2 shows a more complete graph of the Bessel function. For the power series that we have looked at so far, the set of values of x for which the series is convergent has always turned out to be an interval [a finite interval for the geometric series and the series in Example 2, the infinite interval ͑Ϫϱ, ϱ͒ in Example 3, and a collapsed interval ͓0, 0͔ ෇ ͕0͖ in Example 1]. The following theorem, proved in Appendix F, says that this is true in general. J¸ ϱ FIGURE 1 Partial sums of the Bessel function J¸ y 1 10 0 x ͚ c ͑x Ϫ a͒ n n n෇0 Խ y=J¸(x) _10 Theorem For a given power series there are only three possibilities: (i) The series converges only when x ෇ a. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x Ϫ a Ͻ R and diverges if x Ϫ a Ͼ R. 3 Խ Խ The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R ෇ 0 in case (i) and R ෇ ϱ in case (ii). The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In case (i) the interval consists of just a single point a. In case (ii) the interval is ͑Ϫϱ, ϱ͒. In case (iii) note that the inequality x Ϫ a Ͻ R can be rewritten as a Ϫ R Ͻ x Ͻ a ϩ R. When x is an endpoint of the interval, that is, x ෇ a Ϯ R, anything can happen—the series might converge at one or both endpoints or it might diverge at both endpoints. Thus, in case (iii) there are four possibilities for the interval of convergence: Խ FIGURE 2 Խ ͑a Ϫ R, a ϩ R͒ ͑a Ϫ R, a ϩ R͔ ͓a Ϫ R, a ϩ R͒ Խ ͓a Ϫ R, a ϩ R͔ The situation is illustrated in Figure 3. convergence for | x-a |<R a-R FIGURE 3 a+R a divergence for | x-a |>R We summarize here the radius and interval of convergence for each of the examples already considered in this section. Series Radius of convergence Interval of convergence R෇1 ͑Ϫ1, 1͒ n R෇0 ͕0͖ ͑x Ϫ 3͒n n R෇1 ͓2, 4͒ ͑Ϫ1͒nx 2n 2 2n͑n!͒2 R෇ϱ ͑Ϫϱ, ϱ͒ ϱ Geometric series ͚x n n෇0 ϱ Example 1 ͚ n! x n෇0 ϱ Example 2 ͚ n෇1 ϱ Example 3 ͚ n෇0 5E-12(pp 786-795) 788 ❙❙❙❙ 1/18/06 10:37 AM Page 788 CHAPTER 12 INFINITE SEQUENCES AND SERIES In general, the Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint of the interval of convergence, so the endpoints must be checked with some other test. EXAMPLE 4 Find the radius of convergence and interval of convergence of the series ϱ ͑Ϫ3͒n x n sn ϩ 1 ͚ n෇0 SOLUTION Let a n ෇ ͑Ϫ3͒n x n͞sn ϩ 1. Then Ϳ Ϳ Ϳ Ϳ Ϳ ͱ Ϳ anϩ1 ͑Ϫ3͒nϩ1x nϩ1 sn ϩ 1 ෇ ؒ ෇ Ϫ3x an ͑Ϫ3͒nx n sn ϩ 2 ෇3 ͱ 1 ϩ ͑1͞n͒ x l 3 x 1 ϩ ͑2͞n͒ Խ Խ Խ Խ nϩ1 nϩ2 as n l ϱ Խ Խ Խ Խ By the Ratio Test, the given series converges if 3 x Ͻ 1 and diverges if 3 x Ͼ 1. Thus, it converges if x Ͻ 1 and diverges if x Ͼ 1 . This means that the radius of 3 3 convergence is R ෇ 1 . 3 We know the series converges in the interval (Ϫ 1 , 1 ), but we must now test for 3 3 convergence at the endpoints of this interval. If x ෇ Ϫ1 , the series becomes 3 Խ Խ ϱ ͚ n෇0 Խ Խ n ϱ ͑Ϫ3͒n(Ϫ1 ) 1 1 1 1 1 3 ෇ ͚ ෇ ϩ ϩ ϩ ϩ иии sn ϩ 1 s1 s2 s3 s4 n෇0 sn ϩ 1 which diverges. (Use the Integral Test or simply observe that it is a p-series with p ෇ 1 Ͻ 1.) If x ෇ 1 , the series is 2 3 ϱ ͚ n෇0 n ϱ ͑Ϫ3͒n( 1 ) ͑Ϫ1͒n 3 ෇ ͚ sn ϩ 1 n෇0 sn ϩ 1 which converges by the Alternating Series Test. Therefore, the given power series converges when Ϫ1 Ͻ x ഛ 1 , so the interval of convergence is (Ϫ1 , 1 ]. 3 3 3 3 EXAMPLE 5 Find the radius of convergence and interval of convergence of the series ϱ ͚ n෇0 n͑x ϩ 2͒n 3 nϩ1 SOLUTION If a n ෇ n͑x ϩ 2͒n͞3 nϩ1, then Ϳ Ϳ Ϳ ͩ ͪԽ anϩ1 ͑n ϩ 1͒͑x ϩ 2͒nϩ1 3 nϩ1 ෇ ؒ an 3 nϩ2 n͑x ϩ 2͒n ෇ 1ϩ 1 n xϩ2 3 Խ l Ϳ Խx ϩ 2Խ 3 Խ as n l ϱ Խ Խ Using the Ratio Test, we see that the series converges if x ϩ 2 ͞3 Ͻ 1 and it diverges if x ϩ 2 ͞3 Ͼ 1. So it converges if x ϩ 2 Ͻ 3 and diverges if x ϩ 2 Ͼ 3. Thus, the radius of convergence is R ෇ 3. Խ Խ Խ Խ Խ 5E-12(pp 786-795) 1/18/06 10:38 AM Page 789 SECTION 12.8 POWER SERIES Խ ❙❙❙❙ 789 Խ The inequality x ϩ 2 Ͻ 3 can be written as Ϫ5 Ͻ x Ͻ 1, so we test the series at the endpoints Ϫ5 and 1. When x ෇ Ϫ5, the series is ϱ ͚ n෇0 ϱ n͑Ϫ3͒n ෇ 1 ͚ ͑Ϫ1͒nn 3 3 nϩ1 n෇0 which diverges by the Test for Divergence [͑Ϫ1͒nn doesn’t converge to 0]. When x ෇ 1, the series is ϱ ϱ n͑3͒n ෇1 ͚ n 3 3 nϩ1 n෇0 ͚ n෇0 which also diverges by the Test for Divergence. Thus, the series converges only when Ϫ5 Ͻ x Ͻ 1, so the interval of convergence is ͑Ϫ5, 1͒. |||| 12.8 Exercises ϱ 1. What is a power series? 25. 2. (a) What is the radius of convergence of a power series? n෇1 ϱ How do you find it? (b) What is the interval of convergence of a power series? How do you find it? 3–28 27. ϱ 28. Find the radius of convergence and interval of convergence of the series. 3. ͚ n෇1 ϱ 5. ͚ n෇1 ϱ 7. ͚ n෇0 ϱ n x sn 4. ͑Ϫ1͒nϪ1 x n n3 6. xn n! 8. n෇0 ϱ ͚ n෇1 n ϱ n 10. ͚ ͑Ϫ1͒ n n෇2 ϱ 12. 14. 17. n ϱ 19. ͚ n෇1 ϱ 16. 21. ͚ n෇1 ϱ 23. ͚ n෇1 n ■ 18. (a) xn 5nn5 (c) n n (b) ͚ n෇1 bϾ0 ϱ 22. ͚ n෇1 ϱ 24. ͚ n෇1 ■ ■ ■ ■ ͚ c ͑Ϫ4͒ n n n෇0 ϱ ͚c (b) n n (d) n෇0 n ͚ ͑Ϫ1͒ c 9 n n n n෇0 31. If k is a positive integer, find the radius of convergence of x 2n ͑2n͒! the series ϱ ͚ n෇0 n ͑n!͒k n x ͑kn͒! ϱ n ; 32. Graph the first several partial sums sn͑x͒ of the series ͸n෇0 x , together with the sum function f ͑x͒ ෇ 1͑͞1 Ϫ x͒, on a common screen. On what interval do these partial sums appear to be converging to f ͑x͒? ͑Ϫ2͒ ͑x ϩ 3͒n sn n ͑3x Ϫ 2͒n n3 n 33. The function J1 defined by J1͑x͒ ෇ n͑x Ϫ 4͒n n3 ϩ 1 n2xn 2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒ n ϱ ͚ c ͑Ϫ3͒ n ͚c8 n෇0 ϱ 3 ϱ 20. ■ ϱ ͚ c ͑Ϫ2͒ n෇0 ͚ n ͑x Ϫ 5͒ ͚ ■ ϱ xn n3 n n n෇1 ■ when x ෇ 6. What can be said about the convergence or divergence of the following series? ͚ ͑Ϫ1͒ ϱ n ͑x Ϫ 2͒n nn n!͑2x Ϫ 1͒n ■ n 30. Suppose that ͸ϱ cn x n converges when x ෇ Ϫ4 and diverges n෇0 xn n෇0 ͑x ϩ 2͒ n2 n n ͑x Ϫ a͒n, bn ■ ϱ n ͚ ͑Ϫ1͒ 2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒ xn 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ n෇0 n෇0 ͚ sn ͑x Ϫ 1͒ n෇1 x ͑ln n͒n ■ (a) n ϱ xn 4 ln n n n෇0 ϱ ͚ n෇1 ϱ 15. ͚ n෇1 ͑Ϫ2͒n x n 4 sn ϱ 13. ͚n ͑2x ϩ 3͒n n ln n cn 4 is convergent, does it follow that the following series are convergent? n෇1 n n෇2 ϱ ͚ sn x ͚ ͑Ϫ1͒ n 26. n 29. If ͸ ͑Ϫ1͒ x nϩ1 ϱ ͑4x ϩ 1͒n n2 ϱ n෇0 n෇1 n෇1 11. ■ n n ϱ ͚ ͑Ϫ1͒ n 4 x ͚ n෇1 ϱ ϱ 9. ͚ ͚ n෇2 |||| ϱ ͚ ϱ ͚ n෇0 ; ͑Ϫ1͒n x 2nϩ1 n!͑n ϩ 1͒!2 2nϩ1 is called the Bessel function of order 1. (a) Find its domain. (b) Graph the first several partial sums on a common screen. 5E-12(pp 786-795) ❙❙❙❙ 790 CAS 1/18/06 10:39 AM Page 790 CHAPTER 12 INFINITE SEQUENCES AND SERIES A͑x͒ ෇ 1 ϩ CAS ͸ϱ cn x n, where cnϩ4 ෇ cn for all n ജ 0, find the n෇0 interval of convergence of the series and a formula for f ͑x͒. 36. If f ͑x͒ ෇ 34. The function A defined by ; that is, its coefficients are c2n ෇ 1 and c2nϩ1 ෇ 2 for all n ജ 0. Find the interval of convergence of the series and find an explicit formula for f ͑x͒. (c) If your CAS has built-in Bessel functions, graph J1 on the same screen as the partial sums in part (b) and observe how the partial sums approximate J1. x6 x9 x3 ϩ ϩ ϩ иии 2и3 2и3и5и6 2и3и5и6и8и9 is called the Airy function after the English mathematician and astronomer Sir George Airy (1801–1892). (a) Find the domain of the Airy function. (b) Graph the first several partial sums sn͑x͒ on a common screen. (c) If your CAS has built-in Airy functions, graph A on the same screen as the partial sums in part (b) and observe how the partial sums approximate A. Խ Խ n 37. Show that if lim n l ϱ s cn ෇ c, where c 0, then the radius of convergence of the power series ͸ cn x n is R ෇ 1͞c. 38. Suppose that the power series ͸ cn ͑ x Ϫ a͒ n satisfies c n 0 for all n. Show that if lim n l ϱ cn ͞cnϩ1 exists, then it is equal to the radius of convergence of the power series. Խ 39. Suppose the series ͸ cn x n has radius of convergence 2 and the series ͸ dn x n has radius of convergence 3. What is the radius of convergence of the series ͸ ͑cn ϩ dn͒x n ? 40. Suppose that the radius of convergence of the power series 35. A function f is defined by ͸ cn x n is R. What is the radius of convergence of the power series ͸ cn x 2n ? f ͑x͒ ෇ 1 ϩ 2x ϩ x ϩ 2x ϩ x ϩ и и и 2 |||| 12.9 Խ 3 4 Representations of Functions as Power Series In this section we learn how to represent certain types of functions as sums of power series by manipulating geometric series or by differentiating or integrating such a series. You might wonder why we would ever want to express a known function as a sum of infinitely many terms. We will see later that this strategy is useful for integrating functions that don’t have elementary antiderivatives, for solving differential equations, and for approximating functions by polynomials. (Scientists do this to simplify the expressions they deal with; computer scientists do this to represent functions on calculators and computers.) We start with an equation that we have seen before: |||| A geometric illustration of Equation 1 is shown in Figure 1. Because the sum of a series is the limit of the sequence of partial sums, we have 1 ෇ lim sn͑x͒ nlϱ 1Ϫx where 1 ϱ 1 ෇ 1 ϩ x ϩ x2 ϩ x3 ϩ и и и ෇ ͚ xn 1Ϫx n෇0 ԽxԽ Ͻ 1 We first encountered this equation in Example 5 in Section 12.2, where we obtained it by observing that it is a geometric series with a ෇ 1 and r ෇ x. But here our point of view is different. We now regard Equation 1 as expressing the function f ͑x͒ ෇ 1͑͞1 Ϫ x͒ as a sum of a power series. sn͑x͒ ෇ 1 ϩ x ϩ x 2 ϩ и и и ϩ x n s¡¡ y is the n th partial sum. Notice that as n increases, sn͑x͒ becomes a better approximation to f ͑x͒ for Ϫ1 Ͻ x Ͻ 1. sˆ s∞ f s™ FIGURE 1 1 ƒ= and some partial sums 1-x _1 0 1 x 5E-12(pp 786-795) 1/18/06 10:39 AM Page 791 SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 791 EXAMPLE 1 Express 1͑͞1 ϩ x 2 ͒ as the sum of a power series and find the interval of convergence. SOLUTION Replacing x by Ϫx 2 in Equation 1, we have ϱ 1 1 ෇ ෇ ͚ ͑Ϫx 2 ͒n 1 ϩ x2 1 Ϫ ͑Ϫx 2 ͒ n෇0 ෇ ϱ ͚ ͑Ϫ1͒ x n 2n ෇ 1 Ϫ x2 ϩ x4 Ϫ x6 ϩ x8 Ϫ и и и n෇0 Խ Խ Because this is a geometric series, it converges when Ϫx 2 Ͻ 1, that is, x 2 Ͻ 1, or x Ͻ 1. Therefore, the interval of convergence is ͑Ϫ1, 1͒. (Of course, we could have determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here.) Խ Խ EXAMPLE 2 Find a power series representation for 1͑͞x ϩ 2͒. SOLUTION In order to put this function in the form of the left side of Equation 1 we first factor a 2 from the denominator: 1 ෇ 2ϩx ෇ Խ 1 1 ͩ ͪ ͫ ͩ ͪͬ ͚ͩ ͪ ͚ 2 1ϩ 1 2 ϱ n෇0 ෇ x 2 Ϫ 2 1Ϫ Ϫ x 2 n Խ ෇ ϱ n෇0 x 2 ͑Ϫ1͒n n x 2 nϩ1 Խ Խ This series converges when Ϫx͞2 Ͻ 1, that is, x Ͻ 2. So the interval of convergence is ͑Ϫ2, 2͒. EXAMPLE 3 Find a power series representation of x 3͑͞x ϩ 2͒. SOLUTION Since this function is just x 3 times the function in Example 2, all we have to do is to multiply that series by x 3: |||| It’s legitimate to move x 3 across the sigma sign because it doesn’t depend on n. [Use Theorem 12.2.8(i) with c ෇ x 3.] ϱ ϱ x3 1 ͑Ϫ1͒n ͑Ϫ1͒n ෇ x3 ؒ ෇ x 3 ͚ nϩ1 x n ෇ ͚ nϩ1 x nϩ3 xϩ2 xϩ2 n෇0 2 n෇0 2 1 ෇ 1 x 3 Ϫ 1 x 4 ϩ 1 x 5 Ϫ 16 x 6 ϩ и и и 2 4 8 Another way of writing this series is as follows: ϱ x3 ͑Ϫ1͒nϪ1 n ෇ ͚ x xϩ2 2 nϪ2 n෇3 As in Example 2, the interval of convergence is ͑Ϫ2, 2͒. Differentiation and Integration of Power Series The sum of a power series is a function f ͑x͒ ෇ ͸ϱ cn͑x Ϫ a͒n whose domain is the intern෇0 val of convergence of the series. We would like to be able to differentiate and integrate such functions, and the following theorem (which we won’t prove) says that we can do so by differentiating or integrating each individual term in the series, just as we would for a polynomial. This is called term-by-term differentiation and integration. 5E-12(pp 786-795) 792 ❙❙❙❙ 1/18/06 10:40 AM Page 792 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2 Theorem If the power series ͸ cn͑x Ϫ a͒n has radius of convergence R Ͼ 0, then the function f defined by f ͑x͒ ෇ c0 ϩ c1͑x Ϫ a͒ ϩ c2͑x Ϫ a͒2 ϩ и и и ෇ ϱ ͚ c ͑x Ϫ a͒ n n n෇0 is differentiable (and therefore continuous) on the interval ͑a Ϫ R, a ϩ R͒ and (i) f Ј͑x͒ ෇ c1 ϩ 2c2͑x Ϫ a͒ ϩ 3c3͑x Ϫ a͒2 ϩ и и и ෇ ϱ ͚ nc ͑x Ϫ a͒ n nϪ1 n෇1 |||| In part (ii), x c0 dx ෇ c0 x ϩ C1 is written as c0͑x Ϫ a͒ ϩ C, where C ෇ C1 ϩ ac0, so all the terms of the series have the same form. (ii) y f ͑x͒ dx ෇ C ϩ c ͑x Ϫ a͒ ϩ c 0 ෇Cϩ 1 ϱ ͑x Ϫ a͒2 ͑x Ϫ a͒3 ϩ c2 ϩ иии 2 3 ͑x Ϫ a͒nϩ1 nϩ1 ͚c n n෇0 The radii of convergence of the power series in Equations (i) and (ii) are both R. NOTE 1 (iii) ■ Equations (i) and (ii) in Theorem 2 can be rewritten in the form ͚ͫ y ͚ͫ d dx ϱ n෇0 ϱ (iv) n෇0 ͬ ͬ ϱ cn͑x Ϫ a͒n ෇ ͚ n෇0 d ͓cn͑x Ϫ a͒n ͔ dx ϱ cn͑x Ϫ a͒n dx ෇ ͚ y c ͑x Ϫ a͒ dx n n n෇0 We know that, for finite sums, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. Equations (iii) and (iv) assert that the same is true for infinite sums, provided we are dealing with power series. (For other types of series of functions the situation is not as simple; see Exercise 36.) NOTE 2 Although Theorem 2 says that the radius of convergence remains the same when a power series is differentiated or integrated, this does not mean that the interval of convergence remains the same. It may happen that the original series converges at an endpoint, whereas the differentiated series diverges there. (See Exercise 37.) ■ NOTE 3 The idea of differentiating a power series term by term is the basis for a powerful method for solving differential equations. We will discuss this method in Chapter 18. ■ EXAMPLE 4 In Example 3 in Section 12.8 we saw that the Bessel function J0͑x͒ ෇ ϱ ͚ n෇0 ͑Ϫ1͒n x 2n 2 2n͑n!͒2 is defined for all x. Thus, by Theorem 2, J0 is differentiable for all x and its derivative is found by term-by-term differentiation as follows: JЈ͑x͒ ෇ 0 ϱ ͚ n෇0 ϱ d ͑Ϫ1͒nx 2n ͑Ϫ1͒n 2nx 2nϪ1 2n 2 ෇ ͚ dx 2 ͑n!͒ 2 2n͑n!͒2 n෇1 EXAMPLE 5 Express 1͑͞1 Ϫ x͒2 as a power series by differentiating Equation 1. What is the radius of convergence? 5E-12(pp 786-795) 1/18/06 10:41 AM Page 793 SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 793 SOLUTION Differentiating each side of the equation ϱ 1 ෇ 1 ϩ x ϩ x2 ϩ x3 ϩ и и и ෇ ͚ xn 1Ϫx n෇0 we get ϱ 1 ෇ 1 ϩ 2x ϩ 3x 2 ϩ и и и ෇ ͚ nx nϪ1 ͑1 Ϫ x͒2 n෇1 If we wish, we can replace n by n ϩ 1 and write the answer as ϱ 1 ෇ ͚ ͑n ϩ 1͒x n ͑1 Ϫ x͒2 n෇0 According to Theorem 2, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, namely, R ෇ 1. EXAMPLE 6 Find a power series representation for ln͑1 Ϫ x͒ and its radius of convergence. SOLUTION We notice that, except for a factor of Ϫ1, the derivative of this function is 1͑͞1 Ϫ x͒. So we integrate both sides of Equation 1: Ϫln͑1 Ϫ x͒ ෇ y 1 dx ෇ y ͑1 ϩ x ϩ x 2 ϩ и и и͒ dx 1Ϫx ϱ x2 x3 x nϩ1 ϩ ϩ иии ϩ C ෇ ͚ ϩC 2 3 n෇0 n ϩ 1 ෇xϩ ෇ ϱ ͚ n෇1 xn ϩC n ԽxԽ Ͻ 1 To determine the value of C we put x ෇ 0 in this equation and obtain Ϫln͑1 Ϫ 0͒ ෇ C. Thus, C ෇ 0 and ln͑1 Ϫ x͒ ෇ Ϫx Ϫ ϱ x2 x3 xn Ϫ Ϫ и и и ෇ Ϫ͚ 2 3 n෇1 n ԽxԽ Ͻ 1 The radius of convergence is the same as for the original series: R ෇ 1. Notice what happens if we put x ෇ 1 in the result of Example 6. Since ln 1 ෇ Ϫln 2, we 2 2 see that ln 2 ෇ ϱ 1 1 1 1 1 ϩ ϩ ϩ ϩ иии ෇ ͚ 2 8 24 64 n2 n n෇1 EXAMPLE 7 Find a power series representation for f ͑x͒ ෇ tanϪ1x. SOLUTION We observe that f Ј͑x͒ ෇ 1͑͞1 ϩ x 2 ͒ and find the required series by integrating the power series for 1͑͞1 ϩ x 2 ͒ found in Example 1. tanϪ1x ෇ y 1 dx ෇ y ͑1 Ϫ x 2 ϩ x 4 Ϫ x 6 ϩ и и и͒ dx 1 ϩ x2 ෇CϩxϪ x3 x5 x7 ϩ Ϫ ϩ иии 3 5 7 5E-12(pp 786-795) 794 ❙❙❙❙ 1/18/06 10:42 AM Page 794 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| The power series for tanϪ1x obtained in Example 7 is called Gregory’s series after the Scottish mathematician James Gregory (1638–1675), who had anticipated some of Newton’s discoveries. We have shown that Gregory’s series is valid when Ϫ1 Ͻ x Ͻ 1, but it turns out (although it isn’t easy to prove) that it is also valid when x ෇ Ϯ1. Notice that when x ෇ 1 the series becomes 1 1 1 ␲ ෇1Ϫ ϩ Ϫ ϩиии 4 3 5 7 This beautiful result is known as the Leibniz formula for ␲. To find C we put x ෇ 0 and obtain C ෇ tanϪ1 0 ෇ 0. Therefore tanϪ1x ෇ x Ϫ ϱ x3 x5 x7 x 2nϩ1 ϩ Ϫ ϩ и и и ෇ ͚ ͑Ϫ1͒n 3 5 7 2n ϩ 1 n෇0 Since the radius of convergence of the series for 1͑͞1 ϩ x 2 ͒ is 1, the radius of convergence of this series for tanϪ1x is also 1. EXAMPLE 8 (a) Evaluate x ͓1͑͞1 ϩ x 7 ͔͒ dx as a power series. (b) Use part (a) to approximate x00.5 ͓1͑͞1 ϩ x 7 ͔͒ dx correct to within 10Ϫ7. SOLUTION (a) The first step is to express the integrand, 1͑͞1 ϩ x 7 ͒, as the sum of a power series. As in Example 1, we start with Equation 1 and replace x by Ϫx 7: ϱ 1 1 ෇ ෇ ͚ ͑Ϫx 7 ͒n 1 ϩ x7 1 Ϫ ͑Ϫx 7 ͒ n෇0 ϱ ෇ ͚ ͑Ϫ1͒ x n 7n ෇ 1 Ϫ x 7 ϩ x 14 Ϫ и и и n෇0 |||| This example demonstrates one way in which power series representations are useful. Integrating 1͑͞1 ϩ x 7 ͒ by hand is incredibly difficult. Different computer algebra systems return different forms of the answer, but they are all extremely complicated. (If you have a CAS, try it yourself.) The infinite series answer that we obtain in Example 8(a) is actually much easier to deal with than the finite answer provided by a CAS. Now we integrate term by term: 1 y 1ϩx 7 dx ෇ y ϱ ͚ ͑Ϫ1͒ x n 7n dx ෇ C ϩ n෇0 ෇CϩxϪ Խ Խ ϱ ͚ ͑Ϫ1͒ n n෇0 x 7nϩ1 7n ϩ 1 x8 x 15 x 22 ϩ Ϫ ϩ иии 8 15 22 Խ Խ This series converges for Ϫx 7 Ͻ 1, that is, for x Ͻ 1. (b) In applying the Fundamental Theorem of Calculus it doesn’t matter which antiderivative we use, so let’s use the antiderivative from part (a) with C ෇ 0: y 0.5 0 ͫ ͬ 1 x8 x 15 x 22 xϪ ϩ Ϫ ϩ иии 7 dx ෇ 1ϩx 8 15 22 ෇ 1͞2 0 1 1 1 1 ͑Ϫ1͒n Ϫ ϩ иии 8 ϩ 15 Ϫ 22 ϩ и и и ϩ 2 8и2 15 и 2 22 и 2 ͑7n ϩ 1͒2 7nϩ1 This infinite series is the exact value of the definite integral, but since it is an alternating series, we can approximate the sum using the Alternating Series Estimation Theorem. If we stop adding after the term with n ෇ 3, the error is smaller than the term with n ෇ 4: 1 Ϸ 6.4 ϫ 10Ϫ11 29 и 2 29 So we have y 0.5 0 1 1 1 1 1 dx Ϸ Ϫ ϩ Ϫ Ϸ 0.49951374 1 ϩ x7 2 8 ؒ 28 15 ؒ 2 15 22 ؒ 2 22 5E-12(pp 786-795) 1/18/06 10:43 AM Page 795 SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES |||| 12.9 15–18 |||| Find a power series representation for the function and determine the radius of convergence. is 10, what is the radius of convergence of the series ͸ϱ ncn x nϪ1 ? Why? n෇1 2. Suppose you know that the series ͸ ϱ n෇0 15. f ͑x͒ ෇ ln͑5 Ϫ x͒ n bn x converges for Խ x Խ Ͻ 2. What can you say about the following series? Why? ϱ ͚ n෇0 17. f ͑x͒ ෇ bn x nϩ1 nϩ1 ■ ■ x3 ͑x Ϫ 2͒2 ■ ■ ■ ■ 1 1ϩx 4. f ͑x͒ ෇ 3 1 Ϫ x4 19. f ͑x͒ ෇ ln͑3 ϩ x͒ 1 1 Ϫ x3 6. f ͑x͒ ෇ 1 1 ϩ 9x 2 21. f ͑x͒ ෇ ln 7. f ͑x͒ ෇ 1 xϪ5 8. f ͑x͒ ෇ x 4x ϩ 1 11–12 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence. ■ ■ ■ 1ϩx 1Ϫx ■ t 23. y 1Ϫt 25. y ■ ■ ■ 1 x 2 ϩ 25 22. f ͑x͒ ෇ tanϪ1͑2x͒ ■ ■ ■ 8 ■ ■ ■ ■ ■ ■ ■ ln͑1 Ϫ t͒ dt t 24. ■ ■ ■ y 26. dt x Ϫ tan Ϫ1 x dx x3 ■ 3 x2 ϩ x Ϫ 2 y tan Ϫ1 ■ ■ ͑x 2 ͒ dx ■ ■ ■ ■ 27–30 |||| Use a power series to approximate the definite integral to six decimal places. 7x Ϫ 1 12. f ͑x͒ ෇ 3x 2 ϩ 2x Ϫ 1 ■ ■ 23–26 |||| Evaluate the indefinite integral as a power series. What is the radius of convergence? |||| 11. f ͑x͒ ෇ ■ 20. f ͑x͒ ෇ ͩ ͪ x2 10. f ͑x͒ ෇ 3 a Ϫ x3 ■ ■ |||| Find a power series representation for f , and graph f and several partial sums sn͑x͒ on the same screen. What happens as n increases? 5. f ͑x͒ ෇ ■ 18. f ͑x͒ ෇ arctan͑x͞3͒ ■ 3. f ͑x͒ ෇ x 9. f ͑x͒ ෇ 9 ϩ x2 x2 ͑1 Ϫ 2x͒2 16. f ͑x͒ ෇ ; 19–22 3–10 |||| Find a power series representation for the function and determine the interval of convergence. ■ 795 Exercises 1. If the radius of convergence of the power series ͸ϱ cn x n n෇0 ■ ❙❙❙❙ ■ ■ ■ ■ ■ ■ ■ ■ 13. (a) Use differentiation to find a power series representation for f ͑x͒ ෇ 1 ͑1 ϩ x͒2 What is the radius of convergence? (b) Use part (a) to find a power series for f ͑x͒ ෇ 1 ͑1 ϩ x͒3 (c) Use part (b) to find a power series for f ͑x͒ ෇ x2 ͑1 ϩ x͒3 14. (a) Find a power series representation for f ͑x͒ ෇ ln͑1 ϩ x͒. What is the radius of convergence? (b) Use part (a) to find a power series for f ͑x͒ ෇ x ln͑1 ϩ x͒. (c) Use part (a) to find a power series for f ͑x͒ ෇ ln͑x 2 ϩ 1͒. 27. y 29. y ■ 0.2 0 1͞3 0 ■ 1 dx 1 ϩ x5 28. y x 2 tanϪ1͑x 4 ͒ dx 30. y ■ ■ ■ ■ ■ 0.4 0 0.5 0 ■ ln͑1 ϩ x 4͒ dx dx 1 ϩ x6 ■ ■ ■ 31. Use the result of Example 6 to compute ln 1.1 correct to five decimal places. 32. Show that the function f ͑x͒ ෇ ϱ ͚ n෇0 ͑Ϫ1͒n x 2n ͑2n͒! is a solution of the differential equation f Љ͑x͒ ϩ f ͑x͒ ෇ 0 33. (a) Show that J0 (the Bessel function of order 0 given in Example 4) satisfies the differential equation x 2J 0 Љ͑x͒ ϩ xJ0 Ј͑x͒ ϩ x 2J0͑x͒ ෇ 0 (b) Evaluate x01 J0͑x͒ dx correct to three decimal places. ■ 5E-12(pp 796-805) 796 ❙❙❙❙ 1/18/06 10:29 AM Page 796 CHAPTER 12 INFINITE SEQUENCES AND SERIES 34. The Bessel function of order 1 is defined by ϱ ͚ J1͑x͒ ෇ n෇0 ͑Ϫ1͒n x 2nϩ1 n!͑n ϩ 1͒!2 2nϩ1 (a) Show that J1 satisfies the differential equation x 2J1 Љ͑x͒ ϩ x J1 Ј͑x͒ ϩ ͑x 2 Ϫ 1͒J1͑x͒ ෇ 0 the series ϱ ͚ nx ԽxԽ Ͻ 1 nϪ1 n෇1 (b) Find the sum of each of the following series. ϱ ϱ n (i) ͚ nx n, x Ͻ 1 (ii) ͚ n 2 n෇1 n෇1 Խ Խ (c) Find the sum of each of the following series. Ј͑x͒ (b) Show that J0 ෇ ϪJ1͑x͒. ϱ (i) 35. (a) Show that the function f ͑x͒ ෇ 38. (a) Starting with the geometric series ͸ϱ x n, find the sum of n෇0 ͚ n͑n Ϫ 1͒x , Խ x Խ Ͻ 1 n n෇2 ϱ ͚ n෇0 xn n! ϱ (ii) ͚ n෇2 is a solution of the differential equation f Ј͑x͒ ෇ f ͑x͒ ϱ n2 Ϫ n 2n (iii) ͚ n෇1 39. Use the power series for tan n2 2n Ϫ1 x to prove the following expression for ␲ as the sum of an infinite series: ␲ ෇ 2s3 (b) Show that f ͑x͒ ෇ e . x ϱ ͚ n෇0 ͑Ϫ1͒ n ͑2n ϩ 1͒ 3 n 36. Let fn͑x͒ ෇ ͑sin nx͒͞n 2. Show that the series ͸ fn͑x͒ converges 40. (a) By completing the square, show that when x ෇ 2n␲, n an integer. For what values of x does the series ͸ f nЉ͑x͒ converge? y for all values of x but the series of derivatives ͸ f n diverges Ј͑x͒ ϱ ͚ n෇1 xn n2 Find the intervals of convergence for f , f Ј, and f Љ. |||| 12.10 dx ␲ ෇ x2 Ϫ x ϩ 1 3s3 (b) By factoring x 3 ϩ 1 as a sum of cubes, rewrite the integral in part (a). Then express 1͑͞x 3 ϩ 1͒ as the sum of a power series and use it to prove the following formula for ␲ : 37. Let f ͑x͒ ෇ 1͞2 0 ␲෇ 3s3 4 ϱ ͚ n෇0 ͑Ϫ1͒n 8n ͩ 2 1 ϩ 3n ϩ 1 3n ϩ 2 ͪ Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we find such representations? We start by supposing that f is any function that can be represented by a power series 1 f ͑x͒ ෇ c0 ϩ c1͑x Ϫ a͒ ϩ c2͑x Ϫ a͒2 ϩ c3͑x Ϫ a͒3 ϩ c4͑x Ϫ a͒4 ϩ и и и Խx Ϫ aԽ Ͻ R Let’s try to determine what the coefficients cn must be in terms of f . To begin, notice that if we put x ෇ a in Equation 1, then all terms after the first one are 0 and we get f ͑a͒ ෇ c0 By Theorem 12.9.2, we can differentiate the series in Equation 1 term by term: 2 f Ј͑x͒ ෇ c1 ϩ 2c2͑x Ϫ a͒ ϩ 3c3͑x Ϫ a͒2 ϩ 4c4͑x Ϫ a͒3 ϩ и и и and substitution of x ෇ a in Equation 2 gives f Ј͑a͒ ෇ c1 Խx Ϫ aԽ Ͻ R 5E-12(pp 796-805) 1/18/06 10:30 AM Page 797 SECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 797 Now we differentiate both sides of Equation 2 and obtain 3 f Љ͑x͒ ෇ 2c2 ϩ 2 ؒ 3c3͑x Ϫ a͒ ϩ 3 ؒ 4c4͑x Ϫ a͒2 ϩ и и и Խx Ϫ aԽ Ͻ R Again we put x ෇ a in Equation 3. The result is f Љ͑a͒ ෇ 2c2 Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives 4 f ٞ͑x͒ ෇ 2 ؒ 3c3 ϩ 2 ؒ 3 ؒ 4c4͑x Ϫ a͒ ϩ 3 ؒ 4 ؒ 5c5͑x Ϫ a͒2 ϩ и и и Խx Ϫ aԽ Ͻ R and substitution of x ෇ a in Equation 4 gives f ٞ͑a͒ ෇ 2 ؒ 3c3 ෇ 3!c3 By now you can see the pattern. If we continue to differentiate and substitute x ෇ a, we obtain f ͑n͒͑a͒ ෇ 2 ؒ 3 ؒ 4 ؒ и и и ؒ ncn ෇ n!cn Solving this equation for the nth coefficient cn , we get cn ෇ f ͑n͒͑a͒ n! This formula remains valid even for n ෇ 0 if we adopt the conventions that 0! ෇ 1 and f ͑0͒ ෇ f . Thus, we have proved the following theorem. 5 Theorem If f has a power series representation (expansion) at a, that is, if f ͑x͒ ෇ ϱ ͚ c ͑x Ϫ a͒ n n n෇0 Խx Ϫ aԽ Ͻ R then its coefficients are given by the formula cn ෇ f ͑n͒͑a͒ n! Substituting this formula for cn back into the series, we see that if f has a power series expansion at a, then it must be of the following form. 6 f ͑x͒ ෇ ϱ ͚ n෇0 f ͑n͒͑a͒ ͑x Ϫ a͒n n! ෇ f ͑a͒ ϩ f Ј͑a͒ f Љ͑a͒ f ٞ͑a͒ ͑x Ϫ a͒ ϩ ͑x Ϫ a͒2 ϩ ͑x Ϫ a͒3 ϩ и и и 1! 2! 3! 5E-12(pp 796-805) 798 ❙❙❙❙ 1/18/06 10:30 AM Page 798 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| The Taylor series is named after the English mathematician Brook Taylor (1685–1731) and the Maclaurin series is named in honor of the Scottish mathematician Colin Maclaurin (1698–1746) despite the fact that the Maclaurin series is really just a special case of the Taylor series. But the idea of representing particular functions as sums of power series goes back to Newton, and the general Taylor series was known to the Scottish mathematician James Gregory in 1668 and to the Swiss mathematician John Bernoulli in the 1690s. Taylor was apparently unaware of the work of Gregory and Bernoulli when he published his discoveries on series in 1715 in his book Methodus incrementorum directa et inversa. Maclaurin series are named after Colin Maclaurin because he popularized them in his calculus textbook Treatise of Fluxions published in 1742. The series in Equation 6 is called the Taylor series of the function f at a (or about a or centered at a). For the special case a ෇ 0 the Taylor series becomes f ͑x͒ ෇ 7 ϱ ͚ n෇0 f ͑n͒͑0͒ n f Ј͑0͒ f Љ͑0͒ 2 x ෇ f ͑0͒ ϩ xϩ x ϩ иии n! 1! 2! This case arises frequently enough that it is given the special name Maclaurin series. NOTE We have shown that if f can be represented as a power series about a, then f is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. An example of such a function is given in Exercise 62. ■ EXAMPLE 1 Find the Maclaurin series of the function f ͑x͒ ෇ e x and its radius of convergence. SOLUTION If f ͑x͒ ෇ e x, then f ͑n͒͑x͒ ෇ e x, so f ͑n͒͑0͒ ෇ e 0 ෇ 1 for all n. Therefore, the Taylor series for f at 0 (that is, the Maclaurin series) is ϱ ͚ n෇0 ϱ f ͑n͒͑0͒ n xn x x2 x3 x ෇ ͚ ෇1ϩ ϩ ϩ ϩ иии n! 1! 2! 3! n෇0 n! To find the radius of convergence we let a n ෇ x n͞n!. Then Ϳ Ϳ Ϳ Ϳ Խ Խ a nϩ1 x nϩ1 n! x ෇ ؒ n ෇ l 0Ͻ1 an ͑n ϩ 1͒! x nϩ1 so, by the Ratio Test, the series converges for all x and the radius of convergence is R ෇ ϱ. The conclusion we can draw from Theorem 5 and Example 1 is that if e x has a power series expansion at 0, then ex ෇ ϱ ͚ n෇0 xn n! So how can we determine whether e x does have a power series representation? Let’s investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if f has derivatives of all orders, when is it true that ϱ f ͑n͒͑a͒ f ͑x͒ ෇ ͚ ͑x Ϫ a͒n n! n෇0 As with any convergent series, this means that f ͑x͒ is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are n Tn͑x͒ ෇ ͚ i෇0 f ͑i͒͑a͒ ͑x Ϫ a͒i i! ෇ f ͑a͒ ϩ f Ј͑a͒ f Љ͑a͒ f ͑n͒͑a͒ ͑x Ϫ a͒ ϩ ͑x Ϫ a͒2 ϩ и и и ϩ ͑x Ϫ a͒n 1! 2! n! 5E-12(pp 796-805) 1/18/06 10:31 AM Page 799 SECTION 12.10 TAYLOR AND MACLAURIN SERIES y y=T£(x) T1͑x͒ ෇ 1 ϩ x y=T™(x) (0, 1) y=T¡(x) 0 x T2͑x͒ ෇ 1 ϩ x ϩ |||| As n increases, Tn͑x͒ appears to approach e x in Figure 1. This suggests that e x is equal to the sum of its Taylor series. x2 2! T3͑x͒ ෇ 1 ϩ x ϩ x2 x3 ϩ 2! 3! The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1. In general, f ͑x͒ is the sum of its Taylor series if y=T£(x) FIGURE 1 799 Notice that Tn is a polynomial of degree n called the nth-degree Taylor polynomial of f at a. For instance, for the exponential function f ͑x͒ ෇ e x, the result of Example 1 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with n ෇ 1, 2, and 3 are y=´ y=T™(x) ❙❙❙❙ f ͑x͒ ෇ lim Tn͑x͒ nlϱ If we let Rn͑x͒ ෇ f ͑x͒ Ϫ Tn͑x͒ f ͑x͒ ෇ Tn͑x͒ ϩ Rn͑x͒ so that then Rn͑x͒ is called the remainder of the Taylor series. If we can somehow show that lim n l ϱ Rn͑x͒ ෇ 0, then it follows that lim Tn͑x͒ ෇ lim ͓ f ͑x͒ Ϫ Rn͑x͔͒ ෇ f ͑x͒ Ϫ lim Rn͑x͒ ෇ f ͑x͒ nlϱ nlϱ nlϱ We have therefore proved the following. 8 Theorem If f ͑x͒ ෇ Tn͑x͒ ϩ Rn͑x͒, where Tn is the nth-degree Taylor polynomial of f at a and lim Rn͑x͒ ෇ 0 nlϱ for Խ x Ϫ a Խ Ͻ R, then f is equal to the sum of its Taylor series on the interval Խ x Ϫ a Խ Ͻ R. In trying to show that lim n l ϱ Rn͑x͒ ෇ 0 for a specific function f , we usually use the following fact. 9 Խ Խ Խ Խ Taylor’s Inequality If f ͑nϩ1͒͑x͒ ഛ M for x Ϫ a ഛ d , then the remainder Rn͑x͒ of the Taylor series satisfies the inequality M Խ R ͑x͒ Խ ഛ ͑n ϩ 1͒! Խ x Ϫ a Խ n Խ Խ for x Ϫ a ഛ d nϩ1 Խ Խ To see why this is true for n ෇ 1, we assume that f Љ͑x͒ ഛ M . In particular, we have f Љ͑x͒ ഛ M , so for a ഛ x ഛ a ϩ d we have y x a x f Љ͑t͒ dt ഛ y M dt a An antiderivative of f Љ is f Ј, so by Part 2 of the Fundamental Theorem of Calculus, we have f Ј͑x͒ Ϫ f Ј͑a͒ ഛ M͑x Ϫ a͒ or f Ј͑x͒ ഛ f Ј͑a͒ ϩ M͑x Ϫ a͒ 5E-12(pp 796-805) 800 ❙❙❙❙ 1/18/06 10:31 AM Page 800 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| As alternatives to Taylor’s Inequality, we have the following formulas for the remainder term. If f ͑nϩ1͒ is continuous on an interval I and x ʦ I , then 1 x R n͑x͒ ෇ y ͑x Ϫ t͒n f ͑nϩ1͒ ͑t͒ dt n! a This is called the integral form of the remainder term. Another formula, called Lagrange’s form of the remainder term, states that there is a number z between x and a such that f ͑nϩ1͒͑z͒ R n͑x͒ ෇ ͑x Ϫ a͒ nϩ1 ͑n ϩ 1͒! This version is an extension of the Mean Value Theorem (which is the case n ෇ 0). Proofs of these formulas, together with discussions of how to use them to solve the examples of Sections 12.10 and 12.12, are given on the web site y Thus x a x f Ј͑t͒ dt ഛ y ͓ f Ј͑a͒ ϩ M͑t Ϫ a͔͒ dt a f ͑x͒ Ϫ f ͑a͒ ഛ f Ј͑a͒͑x Ϫ a͒ ϩ M f ͑x͒ Ϫ f ͑a͒ Ϫ f Ј͑a͒͑x Ϫ a͒ ഛ ͑x Ϫ a͒2 2 M ͑x Ϫ a͒2 2 But R1͑x͒ ෇ f ͑x͒ Ϫ T1͑x͒ ෇ f ͑x͒ Ϫ f ͑a͒ Ϫ f Ј͑a͒͑x Ϫ a͒. So R1͑x͒ ഛ M ͑x Ϫ a͒2 2 A similar argument, using f Љ͑x͒ ജ ϪM , shows that R1͑x͒ ജ Ϫ M ͑x Ϫ a͒2 2 www.stewartcalculus.com Click on Additional Topics and then on Formulas for the Remainder Term in Taylor series. Խ R ͑x͒ Խ ഛ So 1 M xϪa 2 Խ Խ 2 Although we have assumed that x Ͼ a, similar calculations show that this inequality is also true for x Ͻ a. This proves Taylor’s Inequality for the case where n ෇ 1. The result for any n is proved in a similar way by integrating n ϩ 1 times. (See Exercise 61 for the case n ෇ 2.) NOTE In Section 12.12 we will explore the use of Taylor’s Inequality in approximating functions. Our immediate use of it is in conjunction with Theorem 8. In applying Theorems 8 and 9 it is often helpful to make use of the following fact. ■ lim 10 nlϱ xn ෇0 n! for every real number x This is true because we know from Example 1 that the series and so its nth term approaches 0. ͸ x n͞n! converges for all x EXAMPLE 2 Prove that e x is equal to the sum of its Maclaurin series. SOLUTION If f ͑x͒ ෇ e x, then f ͑nϩ1͒͑x͒ ෇ e x for all n. If d is any positive number and Խ x Խ ഛ d, then Խ f says that ͑nϩ1͒ Խ ͑x͒ ෇ e x ഛ e d. So Taylor’s Inequality, with a ෇ 0 and M ෇ e d, ed Խ R ͑x͒ Խ ഛ ͑n ϩ 1͒! Խ x Խ nϩ1 n Խ Խ for x ഛ d Notice that the same constant M ෇ e d works for every value of n. But, from Equation 10, we have lim nlϱ ed x ͑n ϩ 1͒! Խ Խ nϩ1 ԽxԽ nϩ1 ෇ e d lim nlϱ ͑n ϩ 1͒! ෇0 5E-12(pp 796-805) 1/18/06 10:32 AM Page 801 SECTION 12.10 TAYLOR AND MACLAURIN SERIES Խ ❙❙❙❙ 801 Խ It follows from the Squeeze Theorem that lim n l ϱ Rn͑x͒ ෇ 0 and therefore lim n l ϱ Rn͑x͒ ෇ 0 for all values of x. By Theorem 8, e x is equal to the sum of its Maclaurin series, that is, ex ෇ 11 ϱ ͚ n෇0 |||| In 1748 Leonard Euler used Equation 12 to find the value of e correct to 23 digits. In 2000 Xavier Gourdon and S. Kondo, again using the series in (12), computed e to more than twelve billion decimal places. The special techniques they employed to speed up the computation are explained on the web page xn n! for all x In particular, if we put x ෇ 1 in Equation 11, we obtain the following expression for the number e as a sum of an infinite series: 12 ϱ ͚ e෇ n෇0 1 1 1 1 ෇1ϩ ϩ ϩ ϩ иии n! 1! 2! 3! www.numbers.computation.free.fr EXAMPLE 3 Find the Taylor series for f ͑x͒ ෇ e x at a ෇ 2. SOLUTION We have f ͑n͒͑2͒ ෇ e 2 and so, putting a ෇ 2 in the definition of a Taylor series (6), we get ϱ ͚ n෇0 ϱ f ͑n͒͑2͒ e2 ͑x Ϫ 2͒n ෇ ͚ ͑x Ϫ 2͒n n! n෇0 n! Again it can be verified, as in Example 1, that the radius of convergence is R ෇ ϱ. As in Example 2 we can verify that lim n l ϱ Rn͑x͒ ෇ 0, so ex ෇ 13 ϱ ͚ n෇0 e2 ͑x Ϫ 2͒n n! for all x We have two power series expansions for e x, the Maclaurin series in Equation 11 and the Taylor series in Equation 13. The first is better if we are interested in values of x near 0 and the second is better if x is near 2. EXAMPLE 4 Find the Maclaurin series for sin x and prove that it represents sin x for all x. SOLUTION We arrange our computation in two columns as follows: f ͑x͒ ෇ sin x f ͑0͒ ෇ 0 f Ј͑x͒ ෇ cos x f Ј͑0͒ ෇ 1 f Љ͑x͒ ෇ Ϫsin x f Љ͑0͒ ෇ 0 f ٞ͑x͒ ෇ Ϫcos x f ٞ͑0͒ ෇ Ϫ1 f ͑4͒͑x͒ ෇ sin x f ͑4͒͑0͒ ෇ 0 Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: f Ј͑0͒ f Љ͑0͒ 2 f ٞ͑0͒ 3 f ͑0͒ ϩ xϩ x ϩ x ϩ иии 1! 2! 3! ෇xϪ ϱ x5 x7 x 2nϩ1 x3 ϩ Ϫ ϩ и и и ෇ ͚ ͑Ϫ1͒n 3! 5! 7! ͑2n ϩ 1͒! n෇0 5E-12(pp 796-805) 802 ❙❙❙❙ 1/18/06 10:33 AM Page 802 CHAPTER 12 INFINITE SEQUENCES AND SERIES Խ T1͑x͒ ෇ x T3͑x͒ ෇ x Ϫ x3 3! T5͑x͒ ෇ x Ϫ Խ Since f ͑nϩ1͒͑x͒ is Ϯsin x or Ϯcos x, we know that f ͑nϩ1͒͑x͒ ഛ 1 for all x. So we can take M ෇ 1 in Taylor’s Inequality: |||| Figure 2 shows the graph of sin x together with its Taylor (or Maclaurin) polynomials x3 x5 ϩ 3! 5! Խ 14 Խ Խ M x nϩ1 x nϩ1 ෇ ͑n ϩ 1͒! ͑n ϩ 1͒! Խ Խ Rn͑x͒ ഛ Խ By Equation 10 the right side of this inequality approaches 0 as n l ϱ, so Rn͑x͒ l 0 by the Squeeze Theorem. It follows that Rn͑x͒ l 0 as n l ϱ, so sin x is equal to the sum of its Maclaurin series by Theorem 8. Խ Notice that, as n increases, Tn͑x͒ becomes a better approximation to sin x. Խ y We state the result of Example 4 for future reference. T¡ 1 T∞ y=sin x sin x ෇ x Ϫ 15 0 x 1 ෇ ϱ x 2nϩ1 ͑2n ϩ 1͒! ͚ ͑Ϫ1͒ n n෇0 T£ FIGURE 2 x3 x5 x7 ϩ Ϫ ϩ иии 3! 5! 7! for all x EXAMPLE 5 Find the Maclaurin series for cos x. SOLUTION We could proceed directly as in Example 4 but it’s easier to differentiate the Maclaurin series for sin x given by Equation 15: cos x ෇ d d ͑sin x͒ ෇ dx dx ෇1Ϫ |||| The Maclaurin series for e x, sin x, and cos x that we found in Examples 2, 4, and 5 were discovered, using different methods, by Newton. (There is evidence that the series for sin x and cos x were known to Indian astronomers more than a century before Newton, but this knowledge didn’t spread to the rest of the world.) These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0. ͩ ͪ x3 x5 x7 ϩ Ϫ ϩ иии 3! 5! 7! xϪ 3x 2 5x 4 7x 6 x2 x4 x6 ϩ Ϫ ϩ иии ෇ 1 Ϫ ϩ Ϫ ϩ иии 3! 5! 7! 2! 4! 6! Since the Maclaurin series for sin x converges for all x, Theorem 2 in Section 12.9 tells us that the differentiated series for cos x also converges for all x. Thus 16 cos x ෇ 1 Ϫ ෇ x4 x6 x2 ϩ Ϫ ϩ иии 2! 4! 6! ϱ ͚ ͑Ϫ1͒ n n෇0 x 2n ͑2n͒! for all x EXAMPLE 6 Find the Maclaurin series for the function f ͑x͒ ෇ x cos x. SOLUTION Instead of computing derivatives and substituting in Equation 7, it’s easier to multiply the series for cos x (Equation 16) by x: x cos x ෇ x ϱ ͚ ͑Ϫ1͒ n n෇0 ϱ x 2n x 2nϩ1 ෇ ͚ ͑Ϫ1͒n ͑2n͒! ͑2n͒! n෇0 EXAMPLE 7 Represent f ͑x͒ ෇ sin x as the sum of its Taylor series centered at ␲͞3. 5E-12(pp 796-805) 1/18/06 10:34 AM Page 803 SECTION 12.10 TAYLOR AND MACLAURIN SERIES SOLUTION Arranging our work in columns, we have ͩ ͩ ͩ ͩ ␲ 3 f ͑x͒ ෇ sin x f Ј͑x͒ ෇ cos x |||| We have obtained two different series representations for sin x, the Maclaurin series in Example 4 and the Taylor series in Example 7. It is best to use the Maclaurin series for values of x near 0 and the Taylor series for x near ␲͞3. Notice that the third Taylor polynomial T3 in Figure 3 is a good approximation to sin x near ␲͞3 but not as good near 0. Compare it with the third Maclaurin polynomial T3 in Figure 2, where the opposite is true. y=sin x 0 ␲ 3 fЉ ␲ 3 f ٞ͑x͒ ෇ Ϫcos x fٞ ␲ 3 ͪ ͪ ͪ ͪ ෇ s3 2 ෇ 803 1 2 ෇Ϫ s3 2 ෇Ϫ 1 2 and this pattern repeats indefinitely. Therefore, the Taylor series at ␲͞3 is ͩ ͪ ͩ ͪͩ ͪ ͩ ͪͩ ͪ ͩ ͪ ͩ ͪ fЈ ϩ ෇ x π 3 fЈ f Љ͑x͒ ෇ Ϫsin x ␲ f 3 y f ❙❙❙❙ ␲ 3 1! fЉ ␲ xϪ 3 1 s3 ϩ 2 2 ؒ 1! xϪ ϩ ␲ 3 Ϫ ␲ 3 2! ␲ xϪ 3 s3 2 ؒ 2! xϪ ␲ 3 fٞ 2 ϩ 2 Ϫ ͩ ͪͩ ͪ ͩ ͪ ␲ 3 3! 1 2 ؒ 3! xϪ xϪ ␲ 3 ␲ 3 3 ϩ иии 3 ϩ иии The proof that this series represents sin x for all x is very similar to that in Example 4. [Just replace x by x Ϫ ␲͞3 in (14).] We can write the series in sigma notation if we separate the terms that contain s3 : T£ ϱ sin x ෇ FIGURE 3 ͚ n෇0 ͩ ͪ ͑Ϫ1͒ns3 ␲ xϪ 2͑2n͒! 3 2n ϩ ϱ ͚ n෇0 ͩ ͪ ͑Ϫ1͒n ␲ xϪ 2͑2n ϩ 1͒! 3 2nϩ1 The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 12.9 are indeed the Taylor or Maclaurin series of the given functions because Theorem 5 asserts that, no matter how a power series representation f ͑x͒ ෇ ͸ cn͑x Ϫ a͒n is obtained, it is always true that cn ෇ f ͑n͒͑a͒͞n!. In other words, the coefficients are uniquely determined. We collect in the following table, for future reference, some important Maclaurin series that we have derived in this section and the preceding one. Important Maclaurin series and their intervals of convergence ϱ 1 ෇ ͚ xn ෇ 1 ϩ x ϩ x2 ϩ x3 ϩ и и и 1Ϫx n෇0 ex ෇ ϱ xn x x2 x3 ෇1ϩ ϩ ϩ ϩ иии n! 1! 2! 3! ͚ n෇0 sin x ෇ ϱ n෇0 ͑Ϫϱ, ϱ͒ x 2n x2 x4 x6 ෇1Ϫ ϩ Ϫ ϩ иии ͑2n͒! 2! 4! 6! n ͑Ϫϱ, ϱ͒ ϱ ͚ ͑Ϫ1͒n n෇0 tanϪ1x ෇ ϱ ͚ ͑Ϫ1͒ n n෇0 ͑Ϫϱ, ϱ͒ x3 x5 x7 x 2nϩ1 ෇xϪ ϩ Ϫ ϩ иии ͑2n ϩ 1͒! 3! 5! 7! ͚ ͑Ϫ1͒ cos x ෇ ͑Ϫ1, 1͒ x 2nϩ1 x3 x5 x7 ෇xϪ ϩ Ϫ ϩ иии 2n ϩ 1 3 5 7 ͓Ϫ1, 1͔ 5E-12(pp 796-805) 804 ❙❙❙❙ 1/18/06 10:34 AM Page 804 CHAPTER 12 INFINITE SEQUENCES AND SERIES Module 12.10/12.12 enables you to see how successive Taylor polynomials approach the original function. One reason that Taylor series are important is that they enable us to integrate functions that we couldn’t previously handle. In fact, in the introduction to this chapter we mentioned that Newton often integrated functions by first expressing them as power series and 2 then integrating the series term by term. The function f ͑x͒ ෇ eϪx can’t be integrated by techniques discussed so far because its antiderivative is not an elementary function (see Section 8.5). In the following example we use Newton’s idea to integrate this function. EXAMPLE 8 2 (a) Evaluate x eϪx dx as an infinite series. 2 (b) Evaluate x01 eϪx dx correct to within an error of 0.001. SOLUTION 2 (a) First we find the Maclaurin series for f ͑x͒ ෇ eϪx . Although it’s possible to use the direct method, let’s find it simply by replacing x with Ϫx 2 in the series for e x given in the table of Maclaurin series. Thus, for all values of x, 2 eϪx ෇ ϱ ͚ n෇0 ͑Ϫx 2 ͒n ෇ n! ϱ ͚ ͑Ϫ1͒ n n෇0 x 2n x2 x4 x6 ෇1Ϫ ϩ Ϫ ϩ иии n! 1! 2! 3! Now we integrate term by term: Ϫx 2 ye ͩ dx ෇ y 1 Ϫ ͪ x2 x4 x6 x 2n ϩ Ϫ ϩ и и и ϩ ͑Ϫ1͒n ϩ и и и dx 1! 2! 3! n! ෇CϩxϪ x3 x5 x7 x 2nϩ1 ϩ Ϫ ϩ и и и ϩ ͑Ϫ1͒n ϩ иии 3 ؒ 1! 5 ؒ 2! 7 ؒ 3! ͑2n ϩ 1͒n! 2 This series converges for all x because the original series for eϪx converges for all x. (b) The Evaluation Theorem gives |||| We can take C ෇ 0 in the antiderivative in part (a). y 1 0 Ϫx 2 e ͫ ͬ x3 x5 x7 x9 dx ෇ x Ϫ ϩ Ϫ ϩ Ϫ иии 3 ؒ 1! 5 ؒ 2! 7 ؒ 3! 9 ؒ 4! ෇1Ϫ ϩ 1 3 1 10 Ϫ 1 42 ϩ 1 216 1 0 Ϫ иии 1 1 1 Ϸ 1 Ϫ 1 ϩ 10 Ϫ 42 ϩ 216 Ϸ 0.7475 3 The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 1 1 ෇ Ͻ 0.001 11 ؒ 5! 1320 Another use of Taylor series is illustrated in the next example. The limit could be found with l’Hospital’s Rule, but instead we use a series. EXAMPLE 9 Evaluate lim xl0 ex Ϫ 1 Ϫ x . x2 5E-12(pp 796-805) 1/18/06 10:35 AM Page 805 SECTION 12.10 TAYLOR AND MACLAURIN SERIES SOLUTION Using the Maclaurin series for e x, we have ex Ϫ 1 Ϫ x lim ෇ lim xl0 xl0 x2 |||| Some computer algebra systems compute limits in this way. ͩ 1ϩ xl0 ෇ ͩ 805 ͪ x x2 x3 ϩ ϩ ϩ иии Ϫ 1 Ϫ x 1! 2! 3! x2 x2 x3 x4 ϩ ϩ ϩ иии 2! 3! 4! ෇ lim xl0 x2 ෇ lim ❙❙❙❙ ͪ 1 x x2 x3 ϩ ϩ ϩ ϩ иии 2 3! 4! 5! 1 2 because power series are continuous functions. Multiplication and Division of Power Series If power series are added or subtracted, they behave like polynomials (Theorem 12.2.8 shows this). In fact, as the following example illustrates, they can also be multiplied and divided like polynomials. We find only the first few terms because the calculations for the later terms become tedious and the initial terms are the most important ones. EXAMPLE 10 Find the first three nonzero terms in the Maclaurin series for (a) e x sin x and (b) tan x. SOLUTION (a) Using the Maclaurin series for e x and sin x in the table on page 803, we have ͩ e x sin x ෇ 1 ϩ ͪͩ x x2 x3 ϩ ϩ ϩ иии 1! 2! 3! xϪ ͪ x3 ϩ иии 3! We multiply these expressions, collecting like terms just as for polynomials: 1 ϩ x ϩ 1 x2 ϩ 1 x3 ϩ и и и 2 6 x Ϫ 1 x3 ϩ и и и 6 x ϩ x2 ϩ 1 x3 ϩ 1 x4 ϩ и и и 2 6 Ϫ 1 x3 Ϫ 1 x4 Ϫ и и и 6 6 x ϩ x2 ϩ 1 x3 ϩ и и и 3 Thus e x sin x ෇ x ϩ x 2 ϩ 1 x 3 ϩ и и и 3 (b) Using the Maclaurin series in the table, we have x3 x5 ϩ Ϫ иии sin x 3! 5! tan x ෇ ෇ cos x x2 x4 1Ϫ ϩ Ϫ иии 2! 4! xϪ 5E-12(pp 806-815) ❙❙❙❙ 806 1/18/06 10:18 AM Page 806 CHAPTER 12 INFINITE SEQUENCES AND SERIES We use a procedure like long division: x ϩ 1 x3 ϩ 3 1 1 Ϫ 1 x 2 ϩ 24 x 4 Ϫ и и и͒x Ϫ 1 x 3 ϩ 2 6 x Ϫ 1 x3 ϩ 2 2 15 1 120 1 24 x5 ϩ и и и x3 Ϫ x3 Ϫ 1 30 1 6 x5 ϩ и и и x5 ϩ и и и 2 15 x5 ϩ и и и 1 3 1 3 x5 Ϫ и и и x 5 Ϫ и ии 2 tan x ෇ x ϩ 1 x 3 ϩ 15 x 5 ϩ и и и 3 Thus Although we have not attempted to justify the formal manipulations used in Example 10, they are legitimate. There is a theorem which states that if both f ͑x͒ ෇ ͸ cn x n and t͑x͒ ෇ ͸ bn x n converge for x Ͻ R and the series are multiplied as if they were polynomials, then the resulting series also converges for x Ͻ R and represents f ͑x͒t͑x͒. For division we require b0 0; the resulting series converges for sufficiently small x . Խ Խ |||| 12.10 1. If f ͑x͒ ෇ Խ Խ Խ Խ Exercises ͸ϱ bn͑x Ϫ 5͒ n for all x, write a formula for b 8. n෇0 |||| Find the Taylor series for f ͑x͒ centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn͑x͒ l 0.] 11–18 2. (a) The graph of f is shown. Explain why the series 1.6 Ϫ 0.8͑x Ϫ 1͒ ϩ 0.4͑x Ϫ 1͒2 Ϫ 0.1͑x Ϫ 1͒3 ϩ и и и 11. f ͑x͒ ෇ 1 ϩ x ϩ x 2, a෇2 12. f ͑x͒ ෇ x , a ෇ Ϫ1 13. f ͑x͒ ෇ e x, a෇3 3 is not the Taylor series of f centered at 1. y 14. f ͑x͒ ෇ ln x, f a෇2 15. f ͑x͒ ෇ cos x, 16. f ͑x͒ ෇ sin x, 0 x 1 a ෇ ␲͞2 17. f ͑x͒ ෇ 1͞sx, 1 a෇␲ a෇9 Ϫ2 18. f ͑x͒ ෇ x , ■ (b) Explain why the series ■ ■ a෇1 ■ ■ ■ ■ ■ ■ ■ ■ 19. Prove that the series obtained in Exercise 3 represents cos x for 2.8 ϩ 0.5͑x Ϫ 2͒ ϩ 1.5͑x Ϫ 2͒2 Ϫ 0.1͑x Ϫ 2͒3 ϩ и и и all x. 20. Prove that the series obtained in Exercise 16 represents sin x is not the Taylor series of f centered at 2. for all x. |||| Find the Maclaurin series for f ͑x͒ using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn͑x͒ l 0.] Also find the associated radius of convergence. 21. Prove that the series obtained in Exercise 9 represents sinh x 3–10 for all x. 22. Prove that the series obtained in Exercise 10 represents cosh x for all x. 3. f ͑x͒ ෇ cos x 4. f ͑x͒ ෇ sin 2x 5. f ͑x͒ ෇ ͑1 ϩ x͒Ϫ3 6. f ͑x͒ ෇ ln͑1 ϩ x͒ 23–32 |||| Use a Maclaurin series derived in this section to obtain the Maclaurin series for the given function. 7. f ͑x͒ ෇ e 5x 8. f ͑x͒ ෇ xe x 23. f ͑x͒ ෇ cos ␲ x 24. f ͑x͒ ෇ eϪx͞2 25. f ͑x͒ ෇ x tanϪ1x 26. f ͑x͒ ෇ sin͑x 4 ͒ 27. f ͑x͒ ෇ x 2eϪx 28. f ͑x͒ ෇ x cos 2x 9. f ͑x͒ ෇ sinh x ■ ■ ■ 10. f ͑x͒ ෇ cosh x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-12(pp 806-815) 1/18/06 10:18 AM Page 807 SECTION 12.10 TAYLOR AND MACLAURIN SERIES 29. f ͑x͒ ෇ sin 2x 1 [Hint: Use sin 2x ෇ 2 ͑1 Ϫ cos 2x͒.] xl0 ͭ ͭ ■ sin x x 31. f ͑x͒ ෇ 1 32. f ͑x͒ ෇ ■ ■ if x if x if x ෇ 0 ■ ■ ■ ■ ■ ■ ■ 35. f ͑x͒ ෇ cos͑x ͒ 36. f ͑x͒ ෇ 2 ■ 2 53. y ෇ x ■ ■ ■ ■ |||| n n෇0 57. Evaluate the indefinite integral as an infinite series. ͚ n෇0 ͒ dx 40. ϩ 1 dx 42. y 41. y sx ■ 3 sin x dx x ex Ϫ 1 dx x y ■ ■ ■ ■ ■ ■ ■ ■ ■ |||| Use series to approximate the definite integral to within the indicated accuracy. y 44. y 45. 46. 1 x cos͑ x 3 ͒ dx 0 0.2 0 y 0.1 0 y 0.5 0 ■ 47–49 2 Ϫx 2 x e ■ ■ |||| ■ n ■ ■ ■ ■ ϱ n෇0 ϱ 2nϩ1 ͚ 58. n෇0 ■ ■ ■ ͑Ϫ1͒n␲ 2n 6 2n͑2n͒! 3n 5n n! 27 81 9 ϩ ϩ ϩ иии 2! 3! 4! ■ ͑ln 2͒3 ͑ln 2͒2 Ϫ ϩ иии 2! 3! ■ 2 ■ ■ ■ ■ ■ ■ ■ ■ M xϪa 6 Խ Խ Խ Խ for x Ϫ a ഛ d 3 62. (a) Show that the function defined by ■ ■ ■ ■ ͭ eϪ1͞x f ͑x͒ ෇ 0 ■ Use series to evaluate the limit. x Ϫ tanϪ1x 47. lim xl0 x3 ■ ͚ 56. Խ R ͑x͒ Խ ഛ (five decimal places) ( Խ error Խ Ͻ 0.001) ■ ■ Խ f ٞ͑x͒ Խ ഛ M for Խ x Ϫ a Խ ഛ d, then ( Խ error Խ Ͻ 10Ϫ8) dx ■ 61. Prove Taylor’s Inequality for n ෇ 2, that is, prove that if (three decimal places) ͓tan Ϫ1͑ x 3 ͒ ϩ sin͑x 3͔͒ dx dx s1 ϩ x 3 ■ x 4n n! 60. 1 Ϫ ln 2 ϩ ■ 43–46 43. ■ ͑Ϫ1͒ ␲ 2nϩ1 4 ͑2n ϩ 1͒! 59. 3 ϩ ■ ■ Find the sum of the series. ͚ ͑Ϫ1͒ ϱ y x cos͑x 54. y ෇ e x ln͑1 Ϫ x͒ ■ ϱ 55. five decimal places. 39. ■ 52. y ෇ sec x x sin x ■ 55–60 38. Use the Maclaurin series for sin x to compute sin 3Њ correct to 3 ■ ■ decimal places. |||| ■ tan x Ϫ x x3 51. y ෇ eϪx cos x 37. Use the Maclaurin series for e x to calculate e Ϫ0.2 correct to five 39–42 ■ |||| Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function. ■ ■ ■ 51–54 2 2 ■ ■ 34. f ͑x͒ ෇ eϪx ϩ cos x ■ ■ We found this limit in Example 4 in Section 7.7 using l’Hospital’s Rule three times. Which method do you prefer? 33. f ͑x͒ ෇ s1 ϩ x ■ ■ 0 |||| Find the Maclaurin series of f (by any method) and its radius of convergence. Graph f and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and f ? ■ ■ xl0 ; 33–36 ■ ■ lim 1 6 ■ ■ 50. Use the series in Example 10(b) to evaluate if x ෇ 0 x Ϫ sin x x3 ■ 0 807 sin x Ϫ x ϩ 1 x 3 6 x5 49. lim 30. f ͑x͒ ෇ cos 2x ❙❙❙❙ 1 Ϫ cos x 48. lim x xl0 1 ϩ x Ϫ e ; 2 if x 0 if x ෇ 0 is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin. 5E-12(pp 806-815) 808 ❙❙❙❙ 1/18/06 10:18 AM Page 808 CHAPTER 12 INFINITE SEQUENCES AND SERIES LABORATORY PROJECT CAS An Elusive Limit This project deals with the function f ͑x͒ ෇ sin͑tan x͒ Ϫ tan͑sin x͒ arcsin͑arctan x͒ Ϫ arctan͑arcsin x͒ 1. Use your computer algebra system to evaluate f ͑x͒ for x ෇ 1, 0.1, 0.01, 0.001, and 0.0001. Does it appear that f has a limit as x l 0? 2. Use the CAS to graph f near x ෇ 0. Does it appear that f has a limit as x l 0? 3. Try to evaluate lim x l 0 f ͑x͒ with l’Hospital’s Rule, using the CAS to find derivatives of the numerator and denominator. What do you discover? How many applications of l’Hospital’s Rule are required? 4. Evaluate lim x l 0 f ͑x͒ by using the CAS to find sufficiently many terms in the Taylor series of the numerator and denominator. (Use the command taylor in Maple or Series in Mathematica.) 5. Use the limit command on your CAS to find lim x l 0 f ͑x͒ directly. (Most computer algebra systems use the method of Problem 4 to compute limits.) 6. In view of the answers to Problems 4 and 5, how do you explain the results of Problems 1 and 2? |||| 12.11 The Binomial Series You may be acquainted with the Binomial Theorem, which states that if a and b are any real numbers and k is a positive integer, then ͑a ϩ b͒k ෇ a k ϩ ka kϪ1b ϩ ϩиии ϩ k͑k Ϫ 1͒ kϪ2 2 k͑k Ϫ 1͒͑k Ϫ 2͒ kϪ3 3 a b ϩ a b 2! 3! k͑k Ϫ 1͒͑k Ϫ 2͒ и и и ͑k Ϫ n ϩ 1͒ kϪn n a b n! ϩ и и и ϩ kab kϪ1 ϩ b k The traditional notation for the binomial coefficients is ͩͪ k 0 ෇1 ͩͪ k n ෇ k͑k Ϫ 1͒͑k Ϫ 2͒ и и и ͑k Ϫ n ϩ 1͒ n! n ෇ 1, 2, . . . , k which enables us to write the Binomial Theorem in the abbreviated form k ͑a ϩ b͒k ෇ ͚ n෇0 ͩͪ k kϪn n a b n 5E-12(pp 806-815) 1/18/06 10:19 AM Page 809 SECTION 12.11 THE BINOMIAL SERIES In particular, if we put a ෇ 1 and b ෇ x, we get k ͑1 ϩ x͒k ෇ 1 ͚ n෇0 ❙❙❙❙ 809 ͩͪ k n x n One of Newton’s accomplishments was to extend the Binomial Theorem (Equation 1) to the case in which k is no longer a positive integer. (See the Writing Project on page 812.) In this case the expression for ͑1 ϩ x͒k is no longer a finite sum; it becomes an infinite series. To find this series we compute the Maclaurin series of ͑1 ϩ x͒k in the usual way: f ͑x͒ ෇ ͑1 ϩ x͒k f ͑0͒ ෇ 1 f Ј͑x͒ ෇ k͑1 ϩ x͒kϪ1 f Ј͑0͒ ෇ k f Љ͑x͒ ෇ k͑k Ϫ 1͒͑1 ϩ x͒kϪ2 f Љ͑0͒ ෇ k͑k Ϫ 1͒ f ٞ͑x͒ ෇ k͑k Ϫ 1͒͑k Ϫ 2͒͑1 ϩ x͒kϪ3 . . . ͑n͒ f ͑x͒ ෇ k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒͑1 ϩ x͒kϪn f ٞ͑0͒ ෇ k͑k Ϫ 1͒͑k Ϫ 2͒ . . . ͑n͒ f ͑0͒ ෇ k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒ Therefore, the Maclaurin series of f ͑x͒ ෇ ͑1 ϩ x͒k is ϱ ͚ n෇0 ϱ f ͑n͒͑0͒ n k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒ n x ෇ ͚ x n! n! n෇0 This series is called the binomial series. If its nth term is a n , then Ϳ Ϳ Ϳ anϩ1 k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒͑k Ϫ n͒x nϩ1 n! ෇ ؒ an ͑n ϩ 1͒! k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒x n Խ kϪn ෇ nϩ1 Ϳ ͿԽ Խ k n x ෇ 1 1ϩ n ԽԽ Խ 1Ϫ Խ Խ x l x Ϳ as n l ϱ Խ Խ Խ Խ Thus, by the Ratio Test, the binomial series converges if x Ͻ 1 and diverges if x Ͼ 1 . The following theorem states that ͑1 ϩ x͒k is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term Rn͑x͒ approaches 0, but that turns out to be quite difficult. The proof outlined in Exercise 19 is much easier. 2 Խ Խ The Binomial Series If k is any real number and x Ͻ 1, then ͑1 ϩ x͒k ෇ 1 ϩ kx ϩ ෇ where ͩͪ k n ϱ ͚ n෇0 ෇ ͩͪ k͑k Ϫ 1͒ 2 k͑k Ϫ 1͒͑k Ϫ 2͒ 3 x ϩ x ϩ иии 2! 3! k n x n k͑k Ϫ 1͒ и и и ͑k Ϫ n ϩ 1͒ n! ͑n ജ 1͒ and ͩͪ k 0 ෇1 5E-12(pp 806-815) 810 ❙❙❙❙ 1/18/06 10:19 AM Page 810 CHAPTER 12 INFINITE SEQUENCES AND SERIES Խ Խ Although the binomial series always converges when x Ͻ 1, the question of whether or not it converges at the endpoints, Ϯ1, depends on the value of k. It turns out that the series converges at 1 if Ϫ1 Ͻ k ഛ 0 and at both endpoints if k ജ 0. Notice that if k is a k k positive integer and n Ͼ k, then the expression for ( n ) contains a factor ͑k Ϫ k͒, so ( n ) ϭ 0 for n Ͼ k. This means that the series terminates and reduces to the ordinary Binomial Theorem (Equation 1) when k is a positive integer. As we have seen, the binomial series is just a special case of the Maclaurin series; it occurs so frequently that it is worth remembering. 1 as a power series. ͑1 ϩ x͒2 EXAMPLE 1 Expand SOLUTION We use the binomial series with k ෇ Ϫ2. The binomial coefficient is ͩ ͪ Ϫ2 n ෇ ͑Ϫ2͒͑Ϫ3͒͑Ϫ4͒ и и и ͑Ϫ2 Ϫ n ϩ 1͒ n! ෇ ͑Ϫ1͒n 2 ؒ 3 ؒ 4 ؒ и и и ؒ n͑n ϩ 1͒ ෇ ͑Ϫ1͒n͑n ϩ 1͒ n! Խ Խ and so, when x Ͻ 1, ϱ 1 ෇ ͑1 ϩ x͒Ϫ2 ෇ ͚ ͑1 ϩ x͒2 n෇0 ͩ ͪ Ϫ2 n x n ϱ ͚ ͑Ϫ1͒ ͑n ϩ 1͒x ෇ n n ෇ 1 Ϫ 2x ϩ 3x 2 Ϫ 4 x 3 ϩ и и и n෇0 EXAMPLE 2 Find the Maclaurin series for the function f ͑x͒ ෇ convergence. 1 and its radius of s4 Ϫ x SOLUTION As given, f ͑x͒ is not quite of the form ͑1 ϩ x͒k so we rewrite it as follows: 1 ෇ s4 Ϫ x 1 ͱͩ ͪ ͱ 4 1Ϫ x 4 ෇ 1 1Ϫ 2 x 4 ෇ 1 2 ͩ ͪ 1Ϫ x 4 Ϫ1͞2 1 Using the binomial series with k ෇ Ϫ 2 and with x replaced by Ϫx͞4, we have 1 1 ෇ 2 s4 Ϫ x ෇ 1 2 ͩ ͪ ͚ ͩ ͪͩ ͪ ͫ ͩ ͪͩ ͪ ( )( ) ͩ ͪ 1Ϫ x 4 1ϩ Ϫ Ϫ1͞2 1 2 ϩ иии ϩ ෇ 1 2 ͫ 1ϩ ෇ 1 2 Ϫ ϱ Ϫ1 2 n n෇0 x 4 ϩ Ϫ Ϫ1 Ϫ3 2 2 2! x 4 n Ϫ x 4 2 ϩ (Ϫ 1)(Ϫ 3)(Ϫ 5) и и и (Ϫ 1 Ϫ n ϩ 1) 2 2 2 2 n! ͩ ͪ ͬ (Ϫ 1)(Ϫ 3)(Ϫ 5) 2 2 2 ͩ ͪ Ϫ x 4 3! n Ϫ x 4 3 ϩ иии ͬ 1 1ؒ3 2 1ؒ3ؒ5 3 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ n xϩ x ϩ иии ϩ x ϩ иии 2 x ϩ 3 8 2!8 3!8 n!8 n 5E-12(pp 806-815) 1/18/06 10:19 AM Page 811 SECTION 12.11 THE BINOMIAL SERIES Խ Խ ❙❙❙❙ 811 Խ Խ We know from (2) that this series converges when Ϫx͞4 Ͻ 1, that is, x Ͻ 4, so the radius of convergence is R ෇ 4. y y= 1 1 œ4-x œ„„„„ T£ T™ T¡ |||| A binomial series is a special case of a Taylor series. Figure 1 shows the graphs of the first three Taylor polynomials computed from the answer to Example 2. _4 0 x 4 FIGURE 1 |||| 12.11 1–8 Exercises 15. (a) Expand f ͑x͒ ෇ x͑͞1 Ϫ x͒2 as a power series. Use the binomial series to expand the function as a power series. State the radius of convergence. |||| 1. s1 ϩ x 1 3. ͑2 ϩ x͒3 ; 9–10 n෇1 ■ ■ ■ ■ ■ ■ ■ (b) Use part (a) to find the sum of the series ϱ ͚ n෇1 ■ ■ ■ ■ f ͑x͒ ෇ 1͞s1 ϩ x 3. (b) Use part (a) to evaluate f ͑9͒͑0͒. 19. Use the following steps to prove (2). 3 10. s1 ϩ 4x ■ ■ ■ ■ ■ ■ ■ ■ ■ 11. (a) Use the binomial series to expand 1͞s1 Ϫ x 2. (b) Use part (a) to find the Maclaurin series for sinϪ1x. 12. (a) Use the binomial series to expand 1͞s1 ϩ x 2. (b) Use part (a) to find the Maclaurin series for sinhϪ1x. 13. (a) Expand s1 ϩ x as a power series. f ͑x͒ ෇ s1 ϩ x 2. (b) Use part (a) to evaluate f ͑10͒͑0͒. 18. (a) Use the binomial series to find the Maclaurin series of Use the binomial series to expand the function as a Maclaurin series and to find the first three Taylor polynomials T1, T2, and T3. Graph the function and these Taylor polynomials in the interval of convergence. ■ n2 2n 17. (a) Use the binomial series to find the Maclaurin series of |||| 9. ͑1 ϩ 2x͒3͞4 n 2n 16. (a) Expand f ͑x͒ ෇ ͑x ϩ x 2 ͒͑͞1 Ϫ x͒3 as a power series. 1 6. 5 s32 Ϫ x x2 8. s2 ϩ x x 7. s4 ϩ x 2 ■ ϱ ͚ 4. ͑1 Ϫ x͒2͞3 4 5. s1 Ϫ 8x ■ (b) Use part (a) to find the sum of the series 1 2. ͑1 ϩ x͒4 ■ k (a) Let t͑x͒ ෇ ͸ϱ ( n )x n. Differentiate this series to show that n෇0 kt͑x͒ tЈ͑x͒ ෇ Ϫ1 Ͻ x Ͻ 1 1ϩx (b) Let h͑x͒ ෇ ͑1 ϩ x͒Ϫkt͑x͒ and show that hЈ͑x͒ ෇ 0. (c) Deduce that t͑x͒ ෇ ͑1 ϩ x͒k. 20. In Exercise 53 in Section 11.2 it was shown that the length of the ellipse x ෇ a sin ␪, y ෇ a cos ␪, where a Ͼ b Ͼ 0, is 3 3 (b) Use part (a) to estimate s1.01 correct to four decimal places. 4 14. (a) Expand 1͞s1 ϩ x as a power series. 4 (b) Use part (a) to estimate 1͞s1.1 correct to three decimal places. L ෇ 4a y ␲͞2 0 s1 Ϫ e 2 sin2 ␪ d␪ where e ෇ sa 2 Ϫ b 2 ͞a is the eccentricity of the ellipse. Expand the integrand as a binomial series and use the result of Exercise 44 in Section 8.1 to express L as a series in powers of the eccentricity up to the term in e 6. 5E-12(pp 806-815) 812 ❙❙❙❙ 1/18/06 10:19 AM Page 812 CHAPTER 12 INFINITE SEQUENCES AND SERIES WRITING PROJECT How Newton Discovered the Binomial Series The Binomial Theorem, which gives the expansion of ͑a ϩ b͒k, was known to Chinese mathematicians many centuries before the time of Newton for the case where the exponent k is a positive integer. In 1665, when he was 22, Newton was the first to discover the infinite series expansion of ͑a ϩ b͒k when k is a fractional exponent (positive or negative). He didn’t publish his discovery, but he stated it and gave examples of how to use it in a letter (now called the epistola prior) dated June 13, 1676, that he sent to Henry Oldenburg, secretary of the Royal Society of London, to transmit to Leibniz. When Leibniz replied, he asked how Newton had discovered the binomial series. Newton wrote a second letter, the epistola posterior of October 24, 1676, in which he explained in great detail how he arrived at his discovery by a very indirect route. He was investigating the areas under the curves y ෇ ͑1 Ϫ x 2 ͒n͞2 from 0 to x for n ෇ 0, 1, 2, 3, 4, . . . . These are easy to calculate if n is even. By observing patterns and interpolating, Newton was able to guess the answers for odd values of n. Then he realized he could get the same answers by expressing ͑1 Ϫ x 2 ͒n͞2 as an infinite series. Write a report on Newton’s discovery of the binomial series. Start by giving the statement of the binomial series in Newton’s notation (see the epistola prior on page 285 of [4] or page 402 of [2]). Explain why Newton’s version is equivalent to Theorem 2 on page 809. Then read Newton’s epistola posterior (page 287 in [4] or page 404 in [2]) and explain the patterns that Newton discovered in the areas under the curves y ෇ ͑1 Ϫ x 2 ͒n͞2. Show how he was able to guess the areas under the remaining curves and how he verified his answers. Finally, explain how these discoveries led to the binomial series. The books by Edwards [1] and Katz [3] contain commentaries on Newton’s letters. 1. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), pp. 178–187. 2. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London: MacMillan Press, 1987). 3. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), pp. 463–466. 4. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, N.J.: Princeton University Press, 1969). |||| 12.12 Applications of Taylor Polynomials In this section we explore two types of applications of Taylor polynomials. First we look at how they are used to approximate functions––computer scientists like them because polynomials are the simplest of functions. Then we investigate how physicists and engineers use them in such fields as relativity, optics, blackbody radiation, electric dipoles, the velocity of water waves, and building highways across a desert. Approximating Functions by Polynomials Suppose that f ͑x͒ is equal to the sum of its Taylor series at a: f ͑x͒ ෇ ϱ ͚ n෇0 f ͑n͒͑a͒ ͑x Ϫ a͒n n! 5E-12(pp 806-815) 1/18/06 10:20 AM Page 813 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 813 In Section 12.10 we introduced the notation Tn͑x͒ for the nth partial sum of this series and called it the n th-degree Taylor polynomial of f at a. Thus n Tn͑x͒ ෇ ͚ i෇0 f ͑i͒͑a͒ ͑x Ϫ a͒i i! ෇ f ͑a͒ ϩ f Ј͑a͒ f Љ͑a͒ f ͑n͒͑a͒ ͑x Ϫ a͒ ϩ ͑x Ϫ a͒2 ϩ и и и ϩ ͑x Ϫ a͒n 1! 2! n! Since f is the sum of its Taylor series, we know that Tn͑x͒ l f ͑x͒ as n l ϱ and so Tn can be used as an approximation to f : f ͑x͒ Ϸ Tn͑x͒. Notice that the first-degree Taylor polynomial y T1͑x͒ ෇ f ͑a͒ ϩ f Ј͑a͒͑x Ϫ a͒ y=´ y=T£(x) y=T™(x) y=T™(x) (0, 1) y=T¡(x) 0 x y=T£(x) FIGURE 1 x ෇ 0.2 x ෇ 3.0 T2͑x͒ T4͑x͒ T6͑x͒ T8͑x͒ T10͑x͒ 1.220000 1.221400 1.221403 1.221403 1.221403 8.500000 16.375000 19.412500 20.009152 20.079665 ex 1.221403 20.085537 is the same as the linearization of f at a that we discussed in Section 3.10. Notice also that T1 and its derivative have the same values at a that f and f Ј have. In general, it can be shown that the derivatives of Tn at a agree with those of f up to and including derivatives of order n (see Exercise 36). To illustrate these ideas let’s take another look at the graphs of y ෇ e x and its first few Taylor polynomials, as shown in Figure 1. The graph of T1 is the tangent line to y ෇ e x at ͑0, 1͒; this tangent line is the best linear approximation to e x near ͑0, 1͒. The graph of T2 is the parabola y ෇ 1 ϩ x ϩ x 2͞2, and the graph of T3 is the cubic curve y ෇ 1 ϩ x ϩ x 2͞2 ϩ x 3͞6, which is a closer fit to the exponential curve y ෇ e x than T2. The next Taylor polynomial T4 would be an even better approximation, and so on. The values in the table give a numerical demonstration of the convergence of the Taylor polynomials Tn͑x͒ to the function y ෇ e x. We see that when x ෇ 0.2 the convergence is very rapid, but when x ෇ 3 it is somewhat slower. In fact, the farther x is from 0, the more slowly Tn͑x͒ converges to e x. When using a Taylor polynomial Tn to approximate a function f , we have to ask the questions: How good an approximation is it? How large should we take n to be in order to achieve a desired accuracy? To answer these questions we need to look at the absolute value of the remainder: Խ R ͑x͒ Խ ෇ Խ f ͑x͒ Ϫ T ͑x͒ Խ n n There are three possible methods for estimating the size of the error: Խ Խ 1. If a graphing device is available, we can use it to graph Rn͑x͒ and thereby esti- mate the error. 2. If the series happens to be an alternating series, we can use the Alternating Series Estimation Theorem. 3. In all cases we can use Taylor’s Inequality (Theorem 12.10.9), which says that if f ͑nϩ1͒͑x͒ ഛ M , then Խ Խ M Խ R ͑x͒ Խ ഛ ͑n ϩ 1͒! Խ x Ϫ a Խ n nϩ1 EXAMPLE 1 3 (a) Approximate the function f ͑x͒ ෇ sx by a Taylor polynomial of degree 2 at a ෇ 8. (b) How accurate is this approximation when 7 ഛ x ഛ 9? 5E-12(pp 806-815) 814 ❙❙❙❙ 1/18/06 10:20 AM Page 814 CHAPTER 12 INFINITE SEQUENCES AND SERIES SOLUTION 3 f ͑x͒ ෇ sx ෇ x 1͞3 (a) f ͑8͒ ෇ 2 f Ј͑x͒ ෇ 1 xϪ2͞3 3 1 f Ј͑8͒ ෇ 12 f Љ͑x͒ ෇ Ϫ 2 xϪ5͞3 9 1 f Љ͑8͒ ෇ Ϫ 144 f ٞ͑x͒ ෇ 10 xϪ8͞3 27 Thus, the second-degree Taylor polynomial is f Ј͑8͒ f Љ͑8͒ ͑x Ϫ 8͒ ϩ ͑x Ϫ 8͒2 1! 2! T2͑x͒ ෇ f ͑8͒ ϩ 1 1 ෇ 2 ϩ 12 ͑x Ϫ 8͒ Ϫ 288 ͑x Ϫ 8͒2 The desired approximation is 1 1 3 sx Ϸ T2͑x͒ ෇ 2 ϩ 12 ͑x Ϫ 8͒ Ϫ 288 ͑x Ϫ 8͒2 (b) The Taylor series is not alternating when x Ͻ 8, so we can’t use the Alternating Series Estimation Theorem in this example. But we can use Taylor’s Inequality with n ෇ 2 and a ෇ 8: M Խ R ͑x͒ Խ ഛ 3! Խ x Ϫ 8 Խ 3 2 Խ Խ where f ٞ͑x͒ ഛ M . Because x ജ 7, we have x 8͞3 ജ 7 8͞3 and so 10 1 10 1 ؒ 8͞3 ഛ ؒ 8͞3 Ͻ 0.0021 27 x 27 7 f ٞ͑x͒ ෇ Therefore, we can take M ෇ 0.0021. Also 7 ഛ x ഛ 9, so Ϫ1 ഛ x Ϫ 8 ഛ 1 and x Ϫ 8 ഛ 1. Then Taylor’s Inequality gives Խ 2.5 Խ Խ R ͑x͒ Խ ഛ T™ 2 #„ y= œx 0.0021 0.0021 ؒ 13 ෇ Ͻ 0.0004 3! 6 Thus, if 7 ഛ x ഛ 9, the approximation in part (a) is accurate to within 0.0004. 15 0 Let’s use a graphing device to check the calculation in Example 1. Figure 2 shows that 3 the graphs of y ෇ sx and y ෇ T2͑x͒ are very close to each other when x is near 8. Figure 3 shows the graph of R2͑x͒ computed from the expression FIGURE 2 Խ 0.0003 Խ Խ R ͑x͒ Խ ෇ Խ sx Ϫ T ͑x͒ Խ 3 2 y=|R™(x)| 2 We see from the graph that Խ R ͑x͒ Խ Ͻ 0.0003 2 7 9 0 FIGURE 3 when 7 ഛ x ഛ 9. Thus, the error estimate from graphical methods is slightly better than the error estimate from Taylor’s Inequality in this case. 5E-12(pp 806-815) 1/18/06 10:20 AM Page 815 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 815 EXAMPLE 2 (a) What is the maximum error possible in using the approximation sin x Ϸ x Ϫ x3 x5 ϩ 3! 5! when Ϫ0.3 ഛ x ഛ 0.3? Use this approximation to find sin 12Њ correct to six decimal places. (b) For what values of x is this approximation accurate to within 0.00005? SOLUTION (a) Notice that the Maclaurin series x3 x5 x7 ϩ Ϫ ϩ иии 3! 5! 7! sin x ෇ x Ϫ is alternating for all nonzero values of x, and the successive terms decrease in size because x Ͻ 1, so we can use the Alternating Series Estimation Theorem. The error in approximating sin x by the first three terms of its Maclaurin series is at most Խ Խ Ϳ Ϳ Խ Խ x7 x 7 ෇ 7! 5040 Խ Խ If Ϫ0.3 ഛ x ഛ 0.3, then x ഛ 0.3, so the error is smaller than ͑0.3͒7 Ϸ 4.3 ϫ 10Ϫ8 5040 To find sin 12Њ we first convert to radian measure. ͩ ͪ ͩ ͪ ͩ ͪ ͩ ͪ sin 12Њ ෇ sin Ϸ 12␲ 180 ␲ Ϫ 15 ෇ sin ␲ 15 3 ␲ 15 1 ϩ 3! ␲ 15 5 1 5! Ϸ 0.20791169 Thus, correct to six decimal places, sin 12Њ Ϸ 0.207912. (b) The error will be smaller than 0.00005 if ԽxԽ 7 5040 Ͻ 0.00005 Solving this inequality for x, we get ԽxԽ 7 Ͻ 0.252 Խ x Խ Ͻ ͑0.252͒ 1͞7 or Ϸ 0.821 Խ Խ So the given approximation is accurate to within 0.00005 when x Ͻ 0.82. Module 12.10/12.12 graphically shows the remainders in Taylor polynomial approximations. What if we use Taylor’s Inequality to solve Example 2? Since f ͑7͒͑x͒ ෇ Ϫcos x, we have f ͑7͒͑x͒ ഛ 1 and so 1 R6͑x͒ ഛ x 7 7! Խ Խ Խ Խ Խ Խ So we get the same estimates as with the Alternating Series Estimation Theorem. 5E-12(pp 816-824) 816 ❙❙❙❙ 1/18/06 10:22 AM Page 816 CHAPTER 12 INFINITE SEQUENCES AND SERIES 4.3 ϫ 10–* What about graphical methods? Figure 4 shows the graph of Խ R ͑x͒ Խ ෇ Խ sin x Ϫ ( x Ϫ x ϩ x ) Խ and we see from it that Խ R ͑x͒ Խ Ͻ 4.3 ϫ 10 when Խ x Խ ഛ 0.3. This is the same estimate that we obtained in Example 2. For part (b) we want Խ R ͑x͒ Խ Ͻ 0.00005, so we graph both y ෇ Խ R ͑x͒ Խ and y ෇ 0.00005 in Figure 5. By placing the cursor on the right intersection point we find that the inequality is satisfied when Խ x Խ Ͻ 0.82. Again this is the same esti1 6 6 y=| Rß(x)| 1 120 3 5 Ϫ8 6 6 6 _0.3 0.3 0 mate that we obtained in the solution to Example 2. If we had been asked to approximate sin 72Њ instead of sin 12Њ in Example 2, it would have been wise to use the Taylor polynomials at a ෇ ␲͞3 (instead of a ෇ 0) because they are better approximations to sin x for values of x close to ␲͞3. Notice that 72Њ is close to 60Њ (or ␲͞3 radians) and the derivatives of sin x are easy to compute at ␲͞3. Figure 6 shows the graphs of the Maclaurin polynomial approximations FIGURE 4 0.00006 y=0.00005 y=| Rß(x)| _1 T3͑x͒ ෇ x Ϫ 1 0 T5͑x͒ ෇ x Ϫ x3 x5 ϩ 3! 5! x3 3! T7͑x͒ ෇ x Ϫ T1͑x͒ ෇ x x3 x5 x7 ϩ Ϫ 3! 5! 7! to the sine curve. You can see that as n increases, Tn͑x͒ is a good approximation to sin x on a larger and larger interval. FIGURE 5 y T¡ T∞ x 0 y=sin x T£ FIGURE 6 T¶ One use of the type of calculation done in Examples 1 and 2 occurs in calculators and computers. For instance, when you press the sin or e x key on your calculator, or when a computer programmer uses a subroutine for a trigonometric or exponential or Bessel function, in many machines a polynomial approximation is calculated. The polynomial is often a Taylor polynomial that has been modified so that the error is spread more evenly throughout an interval. Applications to Physics Taylor polynomials are also used frequently in physics. In order to gain insight into an equation, a physicist often simplifies a function by considering only the first two or three terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an approximation to the function. Taylor’s Inequality can then be used to gauge the accuracy of the approximation. The following example shows one way in which this idea is used in special relativity. EXAMPLE 3 In Einstein’s theory of special relativity the mass of an object moving with velocity v is m0 m෇ s1 Ϫ v 2͞c 2 where m0 is the mass of the object when at rest and c is the speed of light. The kinetic 5E-12(pp 816-824) 1/18/06 10:22 AM Page 817 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 817 energy of the object is the difference between its total energy and its energy at rest: K ෇ mc 2 Ϫ m0 c 2 (a) Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K ෇ 1 m0v 2. 2 (b) Use Taylor’s Inequality to estimate the difference in these expressions for K when v ഛ 100 m͞s. Խ Խ SOLUTION (a) Using the expressions given for K and m, we get K ෇ mc 2 Ϫ m0 c 2 ෇ ͫͩ ͪ ෇ m0 c 2 |||| The upper curve in Figure 7 is the graph of the expression for the kinetic energy K of an object with velocity v in special relativity. The lower curve shows the function used for K in classical Newtonian physics. When v is much smaller than the speed of light, the curves are practically identical. ͑1 ϩ x͒Ϫ1͞2 ෇ 1 Ϫ 1 x ϩ 2 c2 Ϫ1 ͬ (Ϫ 1 )(Ϫ 3 ) x 2 ϩ (Ϫ 1 )(Ϫ 3 )(Ϫ 5) x 3 ϩ и и и 2 2 2 2 2 2! ෇1Ϫ xϩ x Ϫ ͫͩ ͩ K ෇ m0 c 2 and 1 2 3 8 1ϩ 2 3! 5 16 x ϩ иии 3 ͪ ͬ 1 v2 3 v4 5 v6 ϩ иии Ϫ 1 2 ϩ 4 ϩ 2 c 8 c 16 c 6 ͪ 1 v2 3 v4 5 v6 ϩ ϩ ϩ иии 2 c2 8 c4 16 c 6 If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get 1 K =2 m¸√ @ c Ϫ1͞2 v2 Խ Խ ෇ m0 c 2 0 1Ϫ With x ෇ Ϫv 2͞c 2, the Maclaurin series for ͑1 ϩ x͒Ϫ1͞2 is most easily computed as a binomial series with k ෇ Ϫ1 . (Notice that x Ͻ 1 because v Ͻ c.) Therefore, we have 2 K K=mc@-m¸c@ m0 c 2 Ϫ m0 c 2 s1 Ϫ v 2͞c 2 ͩ ͪ √ K Ϸ m0 c 2 FIGURE 7 1 v2 2 c2 ෇ 1 m0 v 2 2 (b) If x ෇ Ϫv 2͞c 2, f ͑x͒ ෇ m0 c 2 ͓͑1 ϩ x͒Ϫ1͞2 Ϫ 1͔, and M is a number such that f Љ͑x͒ ഛ M , then we can use Taylor’s Inequality to write Խ Խ M Խ R ͑x͒ Խ ഛ 2! x 1 2 Խ Խ We have f Љ͑x͒ ෇ 3 m0 c 2͑1 ϩ x͒Ϫ5͞2 and we are given that v ഛ 100 m͞s, so 4 3m0 c 2 Խ f Љ͑x͒ Խ ෇ 4͑1 Ϫ v ͞c 2 ͒ 2 5͞2 ഛ 3m0 c 2 4͑1 Ϫ 100 2͞c 2 ͒5͞2 ͑෇ M͒ Thus, with c ෇ 3 ϫ 10 8 m͞s, Խ Խ Խ Խ R1͑x͒ ഛ 1 3m0 c 2 100 4 ؒ ؒ 4 Ͻ ͑4.17 ϫ 10Ϫ10 ͒m0 2 4͑1 Ϫ 100 2͞c 2 ͒5͞2 c So when v ഛ 100 m͞s, the magnitude of the error in using the Newtonian expression for kinetic energy is at most ͑4.2 ϫ 10Ϫ10 ͒m0. 5E-12(pp 816-824) 818 ❙❙❙❙ 1/18/06 10:22 AM Page 818 CHAPTER 12 INFINITE SEQUENCES AND SERIES Another application to physics occurs in optics. Figure 8 is adapted from Optics, 4th ed., by Eugene Hecht (Reading, MA: Addison-Wesley, 2002), page 153. It depicts a wave from the point source S meeting a spherical interface of radius R centered at C. The ray SA is refracted toward P. ¨r A ¨i Lo h V R ˙ ¨t Li S C so P si n¡ FIGURE 8 n™ Refraction at a spherical interface Using Fermat’s principle that light travels so as to minimize the time taken, Hecht derives the equation 1 n1 n2 1 ϩ ෇ ᐉo ᐉi R ͩ n2si n1so Ϫ ᐉi ᐉo ͪ where n1 and n2 are indexes of refraction and ᐉo , ᐉi , so , and si are the distances indicated in Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have |||| Here we use the identity 2 ᐉo ෇ sR 2 ϩ ͑so ϩ R͒2 Ϫ 2R͑so ϩ R͒ cos ␾ ᐉi ෇ sR 2 ϩ ͑si Ϫ R͒2 ϩ 2R͑si Ϫ R͒ cos ␾ cos͑␲ Ϫ ␾͒ ෇ Ϫcos ␾ Because Equation 1 is cumbersome to work with, Gauss, in 1841, simplified it by using the linear approximation cos ␾ Ϸ 1 for small values of ␾. (This amounts to using the Taylor polynomial of degree 1.) Then Equation 1 becomes the following simpler equation [as you are asked to show in Exercise 32(a)]: 3 n1 n2 n2 Ϫ n1 ϩ ෇ so si R The resulting optical theory is known as Gaussian optics, or first-order optics, and has become the basic theoretical tool used to design lenses. A more accurate theory is obtained by approximating cos ␾ by its Taylor polynomial of degree 3 (which is the same as the Taylor polynomial of degree 2). This takes into account rays for which ␾ is not so small, that is, rays that strike the surface at greater distances h above the axis. In Exercise 32(b) you are asked to use this approximation to derive the more accurate equation 4 ͫ ͩ n2 n2 Ϫ n1 n1 n1 ϩ ෇ ϩ h2 so si R 2so 1 1 ϩ so R ͪ 2 ϩ n2 2si ͩ 1 1 Ϫ R si ͪͬ 2 The resulting optical theory is known as third-order optics. Other applications of Taylor polynomials to physics and engineering are explored in Exercises 30, 31, 33, 34, and 35 and in the Applied Project on page 821. 5E-12(pp 816-824) 1/18/06 10:23 AM Page 819 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS |||| 12.12 20. f ͑x͒ ෇ x ln x, f ͑x͒ ෇ cos x centered at a ෇ 0. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x ෇ ␲͞4, ␲͞2, and ␲. (c) Comment on how the Taylor polynomials converge to f ͑x͒. 4. f ͑x͒ ෇ e , a ෇ 1, a ෇ ␲͞6, 6. f ͑x͒ ෇ cos x, ln x , x 9. f ͑x͒ ෇ xe Ϫ2x a ෇ 1, a ෇ 0, , 10. f ͑x͒ ෇ s3 ϩ x , 2 CAS ■ ■ ■ ■ ■ 13–22 ■ ■ 27. sin x Ϸ x Ϫ 28. cos x Ϸ 1 Ϫ n෇2 ■ ■ ■ ■ ■ ■ ■ ■ ■ n ෇ 2, 14. f ͑x͒ ෇ x , a ෇ 1, n ෇ 2, ■ ■ 15. f ͑x͒ ෇ x a ෇ 1, n ෇ 3, 16. f ͑x͒ ෇ cos x, a ෇ ␲͞3, n ෇ 4, 17. f ͑x͒ ෇ tan x, a ෇ 0, 19. f ͑x͒ ෇ e , a ෇ 0, ■ ■ ■ ■ ■ ■ ■ ■ ␳ ͑t͒ ෇ ␳ 20 e ␣ ͑tϪ20͒ Խ 0.8 ഛ x ഛ 1.2 x2 ■ conductivity and is measured in units of ohm-meters (⍀ -m). The resistivity of a given metal depends on the temperature according to the equation ■ 0.9 ഛ x ഛ 1.1 18. f ͑x͒ ෇ ln͑1 ϩ 2x͒, ■ ( Խ error Խ Ͻ 0.005) at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute? 4 ഛ x ഛ 4.2 n ෇ 3, ■ ( Խ error Խ Ͻ 0.01) 30. The resistivity ␳ of a conducting wire is the reciprocal of the ■ a ෇ 4, , ■ 29. A car is moving with speed 20 m͞s and acceleration 2 m͞s2 n ෇ 1, 3, 5, 7, 9 Խ 2͞3 x4 x2 ϩ 2 24 ■ |||| Ϫ2 x3 6 n෇3 (a) Approximate f by a Taylor polynomial with degree n at the number a. (b) Use Taylor’s Inequality to estimate the accuracy of the approximation f ͑x͒ Ϸ Tn͑x͒ when x lies in the given interval. ; (c) Check your result in part (b) by graphing Rn͑x͒ . 13. f ͑x͒ ෇ sx, ■ |||| Use the Alternating Series Estimation Theorem or Taylor’s Inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error. Check your answer graphically. n෇3 ■ ■ ■ ; 27–28 n෇4 n ෇ 2, 4, 6, 8 ■ ■ 26. How many terms of the Maclaurin series for ln͑1 ϩ x͒ do you |||| Use a computer algebra system to find the Taylor polynomials Tn at a ෇ 0 for the given values of n. Then graph these polynomials and f on the same screen. 12. f ͑x͒ ෇ tan x, ■ the Maclaurin series for e x that should be used to estimate e 0.1 to within 0.00001. 11–12 11. f ͑x͒ ෇ sec x, Ϫ1 ഛ x ഛ 1 ■ correct to five decimal places. n෇3 a ෇ 1, n ෇ 5, ■ 25. Use Taylor’s Inequality to determine the number of terms of n෇3 a ෇ 0, ■ need to use to estimate ln 1.4 to within 0.001? a ෇ 2␲͞3, 7. f ͑x͒ ෇ arcsin x, ■ Ϫ1 ഛ x ഛ 1 24. Use the information from Exercise 16 to estimate cos 69Њ n෇3 5. f ͑x͒ ෇ sin x, ■ n ෇ 4, correct to five decimal places. n෇4 a ෇ 2, ■ 0.5 ഛ x ഛ 1.5 23. Use the information from Exercise 5 to estimate sin 35Њ Find the Taylor polynomial Tn͑x͒ for the function f at the number a. Graph f and Tn on the same screen. ■ a ෇ 0, ■ |||| x n ෇ 3, a ෇ 0, 22. f ͑x͒ ෇ sinh 2x, centered at a ෇ 1. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x ෇ 0.9 and 1.3. (c) Comment on how the Taylor polynomials converge to f ͑x͒. 3. f ͑x͒ ෇ ln x, a ෇ 1, 21. f ͑ x͒ ෇ x sin x, ; 2. (a) Find the Taylor polynomials up to degree 3 for f ͑x͒ ෇ 1͞x 8. f ͑x͒ ෇ 819 Exercises ; 1. (a) Find the Taylor polynomials up to degree 6 for ; 3–10 ❙❙❙❙ 0 ഛ x ഛ 2␲͞3 0 ഛ x ഛ ␲͞6 a ෇ 1, n ෇ 3, 0.5 ഛ x ഛ 1.5 n ෇ 3, 0 ഛ x ഛ 0.1 ; where t is the temperature in ЊC. There are tables that list the values of ␣ (called the temperature coefficient) and ␳ 20 (the resistivity at 20ЊC) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for ␳ ͑t͒ by its first- or second-degree Taylor polynomial at t ෇ 20. (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give ␣ ෇ 0.0039͞ЊC and ␳ 20 ෇ 1.7 ϫ 10 Ϫ8 ⍀ -m. Graph the resistivity of copper and the linear and quadratic approximations for Ϫ250ЊC ഛ t ഛ 1000ЊC. 5E-12(pp 816-824) ❙❙❙❙ 820 ; 1/18/06 10:24 AM Page 820 CHAPTER 12 INFINITE SEQUENCES AND SERIES (c) For what values of t does the linear approximation agree with the exponential expression to within one percent? 31. An electric dipole consists of two electric charges of equal magnitude and opposite signs. If the charges are q and Ϫq and are located at a distance d from each other, then the electric field E at the point P in the figure is E෇ By expanding this expression for E as a series in powers of d͞D, show that E is approximately proportional to 1͞D 3 when P is far away from the dipole. 2␲ _q P d D ͱ (b) Notice that all the terms in the series after the first one have coefficients that are at most 1. Use this fact to compare this 4 series with a geometric series and show that q q Ϫ D2 ͑D ϩ d ͒2 q If ␪ 0 is not too large, the approximation T Ϸ 2␲ sL͞t, obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms: L T Ϸ 2␲ (1 ϩ 1 k 2 ) 4 t 32. (a) Derive Equation 3 for Gaussian optics from Equation 1 by approximating cos ␾ in Equation 2 by its first-degree Taylor polynomial. (b) Show that if cos ␾ is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes Equation 4 for third-order optics. [Hint: Use the first two terms in the binomial series for ᐉϪ1 and ᐉϪ1. Also, use ␾ Ϸ sin ␾.] o i ͱ L (1 ϩ 1 k 2 ) ഛ T ഛ 2␲ 4 t (c) Use the inequalities in part (b) to estimate the period of a pendulum with L ෇ 1 meter and ␪ 0 ෇ 10Њ. How does it compare with the estimate T Ϸ 2␲ sL͞t ? What if ␪ 0 ෇ 42Њ ? 35. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the Earth. (a) If R is the radius of the Earth and L is the length of the highway, show that the correction is C ෇ R sec͑L͞R͒ Ϫ R body of water with depth d, as in the figure, then CϷ tL 2␲ d tanh 2␲ L (a) If the water is deep, show that v Ϸ stL͑͞2␲͒. (b) If the water is shallow, use the Maclaurin series for tanh to show that v Ϸ std. (Thus, in shallow water the velocity of a wave tends to be independent of the length of the wave.) (c) Use the Alternating Series Estimation Theorem to show that if L Ͼ 10d, then the estimate v 2 Ϸ td is accurate to within 0.014tL. L2 5L 4 ϩ 2R 24R 3 (c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of Earth to be 6370 km.) C L R R 36. Show that Tn and f have the same derivatives at a up to order n. L 37. In Section 4.9 we considered Newton’s method for approxi- d mating a root r of the equation f ͑x͒ ෇ 0, and from an initial approximation x 1 we obtained successive approximations x 2 , x 3 , . . . , where 34. The period of a pendulum with length L that makes a maxi- mum angle ␪ 0 with the vertical is ͱ T෇4 L t y ␲͞2 0 x nϩ1 ෇ x n Ϫ dx s1 Ϫ k 2 sin 2x where k ෇ sin( 1 ␪ 0 ) and t is the acceleration due to gravity. (In 2 Exercise 40 in Section 8.7 we approximated this integral using Simpson’s Rule.) (a) Expand the integrand as a binomial series and use the result of Exercise 44 in Section 8.1 to show that T ෇ 2␲ L 4 Ϫ 3k 2 t 4 Ϫ 4k 2 (b) Use a Taylor polynomial to show that 33. If a water wave with length L moves with velocity v across a v2 ෇ ͱ ͱͫ L t 1ϩ ͬ 12 2 12 3 2 4 12 3 25 2 6 k ϩ иии 2 k ϩ 2 2 k ϩ 2 24 2 2426 2 f ͑x n͒ f Ј͑x n͒ Use Taylor’s Inequality with n ෇ 1, a ෇ x n, and x ෇ r to show that if f Љ͑x͒ exists on an interval I containing r, x n , and x nϩ1, and f Љ͑x͒ ഛ M , f Ј͑x͒ ജ K for all x ʦ I , then Խ Խ Խ Խx nϩ1 Խ Խ Ϫr ഛ M xn Ϫ r 2K Խ Խ 2 [This means that if x n is accurate to d decimal places, then x nϩ1 is accurate to about 2d decimal places. More precisely, if the error at stage n is at most 10Ϫm, then the error at stage n ϩ 1 is at most ͑M͞2K ͒10Ϫ2m.] 5E-12(pp 816-824) 1/18/06 10:24 AM Page 821 APPLIED PROJECT RADIATION FROM THE STARS ❙❙❙❙ 821 APPLIED PROJECT Radiation from the Stars Any object emits radiation when heated. A blackbody is a system that absorbs all the radiation that falls on it. For instance, a matte black surface or a large cavity with a small hole in its wall (like a blastfurnace) is a blackbody and emits blackbody radiation. Even the radiation from the Sun is close to being blackbody radiation. Proposed in the late 19th century, the Rayleigh-Jeans Law expresses the energy density of blackbody radiation of wavelength ␭ as f ͑␭ ͒ ෇ 8␲ kT ␭4 where ␭ is measured in meters, T is the temperature in kelvins (K), and k is Boltzmann’s constant. The Rayleigh-Jeans Law agrees with experimental measurements for long wavelengths but disagrees drastically for short wavelengths. [The law predicts that f ͑␭͒ l ϱ as ␭ l 0 ϩ but experiments have shown that f ͑␭͒ l 0.] This fact is known as the ultraviolet catastrophe. In 1900 Max Planck found a better model (known now as Planck’s Law) for blackbody radiation: f ͑␭͒ ෇ 8␲ hc␭Ϫ5 e Ϫ1 hc͑͞␭ kT ͒ where ␭ is measured in meters, T is the temperature (in kelvins), and h ෇ Planck’s constant ෇ 6.6262 ϫ 10Ϫ34 Jиs c ෇ speed of light ෇ 2.997925 ϫ 10 8 m͞s k ෇ Boltzmann’s constant ෇ 1.3807 ϫ 10Ϫ23 J͞K 1. Use l’Hospital’s Rule to show that lim f ͑␭͒ ෇ 0 ␭ l 0ϩ and lim f ͑␭͒ ෇ 0 ␭lϱ for Planck’s Law. So this law models blackbody radiation better than the Rayleigh-Jeans Law for short wavelengths. 2. Use a Taylor polynomial to show that, for large wavelengths, Planck’s Law gives approxi- mately the same values as the Rayleigh-Jeans Law. ; 3. Graph f as given by both laws on the same screen and comment on the similarities and differences. Use T ෇ 5700 K (the temperature of the Sun). (You may want to change from meters to the more convenient unit of micrometers: 1 ␮m ෇ 10Ϫ6 m.) 4. Use your graph in Problem 3 to estimate the value of ␭ for which f ͑␭͒ is a maximum under Planck’s Law. ; 5. Investigate how the graph of f changes as T varies. (Use Planck’s Law.) In particular, graph f for the stars Betelgeuse (T ෇ 3400 K), Procyon (T ෇ 6400 K), and Sirius (T ෇ 9200 K) as well as the Sun. How does the total radiation emitted (the area under the curve) vary with T ? Use the graph to comment on why Sirius is known as a blue star and Betelgeuse as a red star. 5E-12(pp 816-824) ❙❙❙❙ 822 |||| 1/18/06 10:24 AM Page 822 CHAPTER 12 INFINITE SEQUENCES AND SERIES 12 Review ■ CONCEPT CHECK 1. (a) What is a convergent sequence? (c) If a series is convergent by the Alternating Series Test, how do you estimate its sum? (b) What is a convergent series? (c) What does lim n l ϱ an ෇ 3 mean? (d) What does ͸ϱ an ෇ 3 mean? n෇1 8. (a) Write the general form of a power series. (b) What is the radius of convergence of a power series? (c) What is the interval of convergence of a power series? 2. (a) What is a bounded sequence? (b) What is a monotonic sequence? (c) What can you say about a bounded monotonic sequence? 3. (a) What is a geometric series? Under what circumstances is it 9. Suppose f ͑x͒ is the sum of a power series with radius of con- vergence R. (a) How do you differentiate f ? What is the radius of convergence of the series for f Ј? (b) How do you integrate f ? What is the radius of convergence of the series for x f ͑x͒ dx? convergent? What is its sum? (b) What is a p-series? Under what circumstances is it convergent? 4. Suppose ͸ an ෇ 3 and sn is the nth partial sum of the series. What is lim n l ϱ an? What is lim n l ϱ sn? 10. (a) Write an expression for the nth-degree Taylor polynomial 5. State the following. (a) (b) (c) (d) (e) (f) (g) of f centered at a. (b) Write an expression for the Taylor series of f centered at a. (c) Write an expression for the Maclaurin series of f . (d) How do you show that f ͑x͒ is equal to the sum of its Taylor series? (e) State Taylor’s Inequality. The Test for Divergence The Integral Test The Comparison Test The Limit Comparison Test The Alternating Series Test The Ratio Test The Root Test 11. Write the Maclaurin series and the interval of convergence for 6. (a) What is an absolutely convergent series? each of the following functions. (a) 1͑͞1 Ϫ x͒ (b) e x (c) sin x (d) cos x (e) tanϪ1x (b) What can you say about such a series? (c) What is a conditionally convergent series? 7. (a) If a series is convergent by the Integral Test, how do you estimate its sum? (b) If a series is convergent by the Comparison Test, how do you estimate its sum? ■ 2. If lim n l ϱ a n ෇ 0, then ͸ a n is convergent. The series ͸ϱ n Ϫsin 1 is convergent. n෇1 3. If lim n l ϱ a n ෇ L, then lim n l ϱ a 2nϩ1 ෇ L. 4. If ͸ cn6 n is convergent, then ͸ cn͑Ϫ2͒n is convergent. 5. 6. 7. If ͸ cn6 n is convergent, then ͸ cn͑Ϫ6͒n is convergent. If ͸ cn x n diverges when x ෇ 6, then it diverges when x ෇ 10. The Ratio Test can be used to determine whether ͸ 1͞n 3 converges. 8. The Ratio Test can be used to determine whether ͸ 1͞n! converges. 9. If 0 ഛ a n ഛ bn and ͸ bn diverges, then ͸ a n diverges. 12. Write the binomial series expansion of ͑1 ϩ x͒ k. What is the radius of convergence of this series? TRUE-FALSE QUIZ Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. ■ ■ ϱ 10. ͚ n෇0 ͑Ϫ1͒n 1 ෇ n! e 11. If Ϫ1 Ͻ ␣ Ͻ 1, then lim n l ϱ ␣ n ෇ 0. 12. If ͸ a n is divergent, then ͸ a n is divergent. Խ Խ 13. If f ͑x͒ ෇ 2x Ϫ x 2 ϩ 3 x 3 Ϫ и и и converges for all x, then 1 f ٞ͑0͒ ෇ 2. 14. If ͕a n ͖ and ͕bn ͖ are divergent, then ͕a n ϩ bn ͖ is divergent. 15. If ͕a n ͖ and ͕bn ͖ are divergent, then ͕a n bn ͖ is divergent. 16. If ͕a n ͖ is decreasing and a n Ͼ 0 for all n, then ͕a n ͖ is convergent. 17. If a n Ͼ 0 and ͸ a n converges, then ͸ ͑Ϫ1͒n a n converges. 18. If a n Ͼ 0 and lim n l ϱ ͑a nϩ1͞a n ͒ Ͻ 1, then lim n l ϱ a n ෇ 0. 5E-12(pp 816-824) 1/18/06 10:25 AM Page 823 CHAPTER 12 REVIEW ■ EXERCISES 1–8 |||| Determine whether the sequence is convergent or divergent. If it is convergent, find its limit. 2 ϩ n3 1. a n ෇ 1 ϩ 2n 3 27–31 |||| ϱ ͚ n෇1 ϱ 29. 2 Find the sum of the series. 5 28. n ͚ ͓tan ͑n ϩ 1͒ Ϫ tanϪ1n͔ ln n 6. a n ෇ sn 30. 7. ͕͑1 ϩ 3͞n͒4n ͖ 8. ͕͑Ϫ10͒n͞n!͖ 31. 1 Ϫ e ϩ ■ n෇1 ϱ ■ ■ ■ ■ ■ ͚ n෇0 ■ ■ ■ 9. A sequence is defined recursively by the equations a 1 ෇ 1, ϱ n෇1 ϱ 13. ͚ n෇1 ϱ 15. ͚ n෇2 ϱ 17. ͚ n෇1 ϱ 19. ͚ n෇1 ϱ 20. ͚ n෇1 ϱ 22. ͚ n෇1 ■ ■ ϱ n n ϩ1 12. ͚ n෇1 ϱ 3 n 5n 14. 1 nsln n 16. cos 3n 1 ϩ ͑1.2͒n 18. ͚ n෇1 ϱ ͚ n෇1 ϱ ͚ n෇1 n 3n ϩ 1 ͪ ϱ n 2n ͑1 ϩ 2n 2 ͒n ϱ 38. (a) Show that the series ͚ ͑Ϫ1͒ nϪ1 the series n෇1 ■ ■ ■ ϱ 24. n෇1 ϱ 25. ͚ ■ ■ n෇1 ■ ■ ■ ͚ ͑Ϫ1͒ ͚ n෇1 ■ ■ ■ ■ 41. 2 n͑x Ϫ 2͒n ͑n ϩ 2͒! 43. n ϱ 42. ͚ n෇1 ■ ■ ■ ■ ϱ xn n 25 n ͚ ͑Ϫ1͒ n෇1 ■ nϩ1 an n |||| Find the radius of convergence and interval of convergence of the series. 40. ͑Ϫ1͒nsn ln n ͩ ͪ 40–43 ϱ n Ϫ3 nϪ1 ϱ 26. ■ nn is convergent. ͑2n͒! is also absolutely convergent. ■ n෇1 ͑Ϫ1͒n͑n ϩ 1͒3 n 2 2nϩ1 ͑Ϫ1͒nϩ1 correct to four n5 39. Prove that if the series ͸ϱ an is absolutely convergent, then n෇1 sn nϩ1 ϱ n Ϫ1͞3 ͚ n෇1 ͚ nϪ1 ■ the series ͸ϱ ͑2 ϩ 5 n ͒Ϫ1. Estimate the error involved in this n෇1 approximation. |||| Determine whether the series is conditionally convergent, absolutely convergent, or divergent. ͚ ͑Ϫ1͒ ■ 37. Use the sum of the first eight terms to approximate the sum of 23–26 23. ͚ n෇1 sn ϩ 1 Ϫ sn Ϫ 1 n ■ ■ the error in using it as an approximation to the sum of the series. (b) Find the sum of this series correct to five decimal places. ͑Ϫ1͒ sn ϩ 1 n෇1 ■ ■ 36. (a) Find the partial sum s5 of the series ͸ϱ 1͞n 6 and estimate n෇1 ϱ ■ ■ nn ෇ 0. (b) Deduce that lim n l ϱ ͑2n͒! 21. ■ ■ ϱ 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒ 5 nn! ͑Ϫ5͒2n n 29n ■ 35. Find the sum of the series n ͩ ■ decimal places. n2 ϩ 1 n3 ϩ 1 ln ■ 34. For what values of x does the series ͸ϱ ͑ln x͒n converge? n෇1 Determine whether the series is convergent or divergent. 3 ■ 1 est value of N that corresponds to ␧ ෇ 0.1 in the precise definition of a limit. ͚ ■ 33. Show that cosh x ജ 1 ϩ 2 x 2 for all x. 4 Ϫn ; 10. Show that lim n l ϱ n e ෇ 0 and use a graph to find the small- 11. e3 e4 e2 Ϫ ϩ Ϫ иии 2! 3! 4! 32. Express the repeating decimal 4.17326326326 . . . as a fraction. a nϩ1 ෇ 1 ͑a n ϩ 4͒. Show that ͕a n ͖ is increasing and a n Ͻ 2 for 3 all n. Deduce that ͕a n ͖ is convergent and find its limit. |||| ͑Ϫ1͒n ␲ n 3 2n ͑2n͒! ■ ■ 11–22 1 n͑n ϩ 3͒ Ϫ1 4. a n ෇ cos͑n␲͞2͒ ■ ͚ n෇1 n sin n 5. a n ෇ 2 n ϩ1 ■ ϱ 2nϩ1 n3 3. a n ෇ 1 ϩ n2 ■ 823 ■ 27. 9 nϩ1 2. a n ෇ 10 n ❙❙❙❙ ■ ͚ n෇1 ϱ ͚ n෇0 ■ ■ ■ ■ ■ ͑x ϩ 2͒n n4 n 2 n͑x Ϫ 3͒n sn ϩ 3 ■ ■ ■ ■ 5E-12(pp 816-824) 824 ❙❙❙❙ 1/18/06 10:26 AM Page 824 CHAPTER 12 INFINITE SEQUENCES AND SERIES 57. f ͑x͒ ෇ sx, 44. Find the radius of convergence of the series ϱ ͚ n෇1 ͑2n͒! n x ͑n!͒2 a ෇ 1, 58. f ͑x͒ ෇ sec x, ■ 45. Find the Taylor series of f ͑x͒ ෇ sin x at a ෇ ␲͞6. ■ ■ ■ lim xl0 |||| Find the Maclaurin series for f and its radius of convergence. You may use either the direct method (definition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for e x, sin x, and tanϪ1x. 51. f ͑x͒ ෇ sin͑x 4 ͒ F෇ 52. f ͑x͒ ෇ 10 x 4 53. f ͑x͒ ෇ 1͞s16 Ϫ x 54. f ͑x͒ ෇ ͑1 Ϫ 3x͒Ϫ5 ■ 55. Evaluate y ■ ■ ■ ■ ■ ■ ■ ; ■ ■ ex dx as an infinite series. x 1 56. Use series to approximate x0 s1 ϩ x 4 dx correct to two decimal places. 57–58 ■ ■ ■ mtR 2 ͑R ϩ h͒2 where R is the radius of the Earth and t is the acceleration due to gravity. (a) Express F as a series in powers of h͞R. (b) Observe that if we approximate F by the first term in the series, we get the expression F Ϸ mt that is usually used when h is much smaller than R. Use the Alternating Series Estimation Theorem to estimate the range of values of h for which the approximation F Ϸ mt is accurate to within 1%. (Use R ෇ 6400 km.) 61. Suppose that f ͑x͒ ෇ ͸ϱ cn x n for all x. n෇0 c0 ෇ c2 ෇ c4 ෇ и и и ෇ 0 |||| (a) Approximate f by a Taylor polynomial with degree n at the number a. ; (b) Graph f and Tn on a common screen. (c) Use Taylor’s Inequality to estimate the accuracy of the approximation f ͑x͒ Ϸ Tn͑x͒ when x lies in the given interval. ; (d) Check your result in part (c) by graphing Rn͑x͒ . Խ ■ sin x Ϫ x x3 (a) If f is an odd function, show that Խ ■ above the surface of the Earth is 50. f ͑x͒ ෇ xe 2x ■ 0 ഛ x ഛ ␲͞6 ■ 60. The force due to gravity on an object with mass m at a height h 48. f ͑x͒ ෇ tanϪ1͑x 2 ͒ 49. f ͑x͒ ෇ ln͑1 Ϫ x͒ ■ ■ 59. Use series to evaluate the following limit. 47–54 x2 1ϩx 0.9 ഛ x ഛ 1.1 a ෇ 0, n ෇ 2, ■ 46. Find the Taylor series of f ͑x͒ ෇ cos x at a ෇ ␲͞3. 47. f ͑x͒ ෇ n ෇ 3, (b) If f is an even function, show that c1 ෇ c3 ෇ c5 ෇ и и и ෇ 0 2 62. If f ͑x͒ ෇ e x , show that f ͑2n͒͑0͒ ෇ ͑2n͒! . n! 5E-12(pp 825-827) 1/18/06 10:27 AM Page 825 PROBLEMS PLUS 1. If f ͑x͒ ෇ sin͑x 3 ͒, find f ͑15͒͑0͒. 2. A function f is defined by f ͑x͒ ෇ lim nlϱ x 2n Ϫ 1 x 2n ϩ 1 Where is f continuous? 3. (a) Show that tan 2 x ෇ cot 2 x Ϫ 2 cot x. 1 1 (b) Find the sum of the series ϱ ͚ n෇1 P¢ 4 P∞ FIGURE FOR PROBLEM 4 Խ Խ 4. Let ͕Pn ͖ be a sequence of points determined as in the figure. Thus AP1 ෇ 1, P£ 2 8 1 x tan n 2n 2 P™ 1 A 1 P¡ ԽP P Խ ෇ 2 n nϩ1 , and angle APn Pnϩ1 is a right angle. Find lim n l ϱ ЄPn APnϩ1 . nϪ1 5. To construct the snowflake curve, start with an equilateral triangle with sides of length 1. Step 1 in the construction is to divide each side into three equal parts, construct an equilateral triangle on the middle part, and then delete the middle part (see the figure). Step 2 is to repeat Step 1 for each side of the resulting polygon. This process is repeated at each succeeding step. The snowflake curve is the curve that results from repeating this process indefinitely. (a) Let sn , ln , and pn represent the number of sides, the length of a side, and the total length of the n th approximating curve (the curve obtained after Step n of the construction), respectively. Find formulas for sn , ln , and pn . (b) Show that pn l ϱ as n l ϱ. (c) Sum an infinite series to find the area enclosed by the snowflake curve. Parts (b) and (c) show that the snowflake curve is infinitely long but encloses only a finite area. 1 2 3 6. Find the sum of the series 1ϩ 1 1 1 1 1 1 1 ϩ ϩ ϩ ϩ ϩ ϩ ϩ иии 2 3 4 6 8 9 12 where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s. 7. (a) Show that for xy Ϫ1, arctan x Ϫ arctan y ෇ arctan xϪy 1 ϩ xy if the left side lies between Ϫ␲͞2 and ␲͞2. (b) Show that 120 1 arctan 119 Ϫ arctan 239 ෇ ␲ 4 (c) Deduce the following formula of John Machin (1680–1751): 1 4 arctan 1 Ϫ arctan 239 ෇ 5 ␲ 4 825 5E-12(pp 825-827) 1/18/06 10:27 AM Page 826 (d) Use the Maclaurin series for arctan to show that 0.197395560 Ͻ arctan 5 Ͻ 0.197395562 1 (e) Show that 0.004184075 Ͻ arctan 239 Ͻ 0.004184077 1 (f) Deduce that, correct to seven decimal places, ␲ Ϸ 3.1415927 Machin used this method in 1706 to find ␲ correct to 100 decimal places. Recently, with the aid of computers, the value of ␲ has been computed to increasingly greater accuracy. In 1999, Takahashi and Kanada, using methods of Borwein and Brent͞Salamin, calculated the value of ␲ to 206,158,430,000 decimal places! 8. (a) Prove a formula similar to the one in Problem 7(a) but involving arccot instead of arctan. (b) Find the sum of the series ϱ ͚ arccot͑n 2 ϩ n ϩ 1͒ n෇0 9. Find the interval of convergence of ͸ϱ n 3x n and find its sum. n෇1 10. If a 0 ϩ a 1 ϩ a 2 ϩ и и и ϩ a k ෇ 0, show that lim (a0 sn ϩ a1 sn ϩ 1 ϩ a2 sn ϩ 2 ϩ и и и ϩ ak sn ϩ k ) ෇ 0 nlϱ If you don’t see how to prove this, try the problem-solving strategy of using analogy (see page 58). Try the special cases k ෇ 1 and k ෇ 2 first. If you can see how to prove the assertion for these cases, then you will probably see how to prove it in general. ϱ 11. Find the sum of the series ͚ ln n෇2 ͩ 1Ϫ ͪ 1 . n2 12. Suppose you have a large supply of books, all the same size, and you stack them at the edge 1 1 6 8 FIGURE FOR PROBLEM 12 1 4 1 2 of a table, with each book extending farther beyond the edge of the table than the one beneath it. Show that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough. Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends one-sixth of its length beyond the fourth, and so on. (Try it yourself with a deck of cards.) Consider centers of mass. u෇1ϩ x6 x9 x3 ϩ ϩ ϩ иии 3! 6! 9! v෇xϩ 13. Let x4 x7 x 10 ϩ ϩ ϩ иии 4! 7! 10! w෇ x2 x5 x8 ϩ ϩ ϩ иии 2! 5! 8! Show that u 3 ϩ v 3 ϩ w 3 Ϫ 3u vw ෇ 1. 14. If p Ͼ 1, evaluate the expression 1 1 1 ϩ p ϩ p ϩ иии 2p 3 4 1 1 1 1 Ϫ p ϩ p Ϫ p ϩ иии 2 3 4 1ϩ 826 5E-12(pp 825-827) 1/18/06 10:28 AM Page 827 15. Suppose that circles of equal diameter are packed tightly in n rows inside an equilateral tri- angle. (The figure illustrates the case n ෇ 4.) If A is the area of the triangle and An is the total area occupied by the n rows of circles, show that lim nlϱ ␲ An ෇ A 2 s3 16. A sequence ͕a n ͖ is defined recursively by the equations a0 ෇ a1 ෇ 1 n͑n Ϫ 1͒a n ෇ ͑n Ϫ 1͒͑n Ϫ 2͒a nϪ1 Ϫ ͑n Ϫ 3͒a nϪ2 Find the sum of the series ͸ϱ an. n෇0 FIGURE FOR PROBLEM 15 17. Taking the value of x x at 0 to be 1 and integrating a series term-by-term, show that y 1 0 P¡ P∞ P™ P˜ Pˆ Pß n෇1 ͑Ϫ1͒nϪ1 nn 18. Starting with the vertices P1͑0, 1͒, P2͑1, 1͒, P3͑1, 0͒, P4͑0, 0͒ of a square, we construct further points as shown in the figure: P5 is the midpoint of P1 P2, P6 is the midpoint of P2 P3, P7 is the midpoint of P3 P4, and so on. The polygonal spiral path P1 P2 P3 P4 P5 P6 P7 . . . approaches a point P inside the square. (a) If the coordinates of Pn are ͑x n, yn ͒, show that 1 x n ϩ x nϩ1 ϩ x nϩ2 ϩ x nϩ3 ෇ 2 and find a 2 similar equation for the y-coordinates. (b) Find the coordinates of P. 19. If f ͑x͒ ෇ P¡¸ ϱ ͚ x x dx ෇ ͸ϱ cm x m has positive radius of convergence and e f ͑x͒ ෇ ͸ϱ dn x n, show that m෇0 n෇0 n P¢ P¶ ndn ෇ P£ ͚ ic d i nജ1 nϪi i෇1 FIGURE FOR PROBLEM 18 20. Right-angled triangles are constructed as in the figure. Each triangle has height 1 and its base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes indefinitely many turns around P by showing that ͸ ␪ n is a divergent series. 1 1 1 1 ¨£ ¨™ ¨¡ 1 P 21. Consider the series whose terms are the reciprocals of the positive integers that can be written in base 10 notation without using the digit 0. Show that this series is convergent and the sum is less than 90. 22. (a) Show that the Maclaurin series of the function f ͑x͒ ෇ x 1 Ϫ x Ϫ x2 ϱ is ͚ fn x n n෇1 where fn is the nth Fibonacci number, that is, f1 ෇ 1, f2 ෇ 1, and fn ෇ fnϪ1 ϩ fnϪ2 for n ജ 3. [Hint: Write x͑͞1 Ϫ x Ϫ x 2͒ ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ . . . and multiply both sides of this equation by 1 Ϫ x Ϫ x 2.] (b) By writing f ͑x͒ as a sum of partial fractions and thereby obtaining the Maclaurin series in a different way, find an explicit formula for the nth Fibonacci number. 827 ...
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This note was uploaded on 02/08/2010 for the course M 340L taught by Professor Lay during the Spring '10 term at École Normale Supérieure.

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