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Unformatted text preview: 5E12(pp 736745) 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. I nfinite Sequences and Series 5E12(pp 736745) 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
to integrate such functions as e x2 . (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 n 1 .
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 n for the value of the function at the
number n.
NOTATION ■ The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by
an or an n1 EXAMPLE 1 Some sequences can be deﬁned by giving a formula for the n th 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
n 1
1nn
3n an
n1 1 an n
n 1234
n
, , , ,...,
,...
2345
n1 1
1nn
3n 1 23
,,
39 45
,
,...,
27 81 1nn
3n 1 ,... 737 5E12(pp 736745) 738 ❙❙❙❙ 1/18/06 10:08 AM Page 738 CHAPTER 12 INFINITE SEQUENCES AND SERIES (c)
(d) {sn
cos 3 }n sn 3, n cos n
,n
6 {0, 1, s2, s3, . . . , sn 3 an n
6 an 3 0 n0 1, 3, . . .} n
s3 1
, , 0, . . . , cos
,...
22
6 EXAMPLE 2 Find a formula for the general term a n of the sequence 3
,
5 4
5
,
,
25 125 6
7
,
,...
625 3125 assuming that the pattern of the ﬁrst few terms continues.
SOLUTION We are given that a1 3
5 4
25 a2 a3 5
125 6
625 a4 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 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 1. In Example 1(b) the factor 1 n meant we
started with a negative term. Here we want to start with a positive term and so we use
1 n 1 or 1 n 1. Therefore,
an 1 n1 n 2
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 fn 1 fn 2 n 3 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 13thcentury 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 n n 1 , 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 ... 5E12(pp 736745) 1/18/06 10:08 AM Page 739 S ECTION 12.1 SEQUENCES an From Figure 1 or 2 it appears that the terms of the sequence a n
approaching 1 as n becomes large. In fact, the difference 1 1 7 a¶= 8
0 n 1234567 n
n nn ❙❙❙❙ 739 1 are 1
1 n 1 can be made as small as we like by taking n sufﬁciently large. We indicate this by writing FIGURE 2 n lim n nl 1 1 In general, the notation
lim a n nl L 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 nl L or a n l L as n l if we can make the terms a n as close to L as we like by taking n sufﬁciently large.
If lim n l 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 nl L or a n l L as n l  Compare this deﬁnition with Deﬁnition 4.4.5. if for every 0 there is a corresponding integer N such that
an L whenever n N 5E12(pp 736745) 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
,L
is chosen, there exists an
N such that all terms of the sequence from a N 1 onward must lie in that interval.
a¡
F IGURE 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
and y L
if n 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∑
0 FIGURE 5 1234 n N Comparison of Deﬁnition 2 and Deﬁnition 4.4.5 shows that the only difference between
lim n l a n L and lim x l f x
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 f x lim n l an L and f n a n when n is an integer, then L. y y=ƒ L FIGURE 6 0 x 1234 In particular, since we know that limx l
we have
1
lim
0
4
nl nr 1 xr
if r 0 when r 0 (Theorem 4.4.4), 0 If a n becomes large as n becomes large, we use the notation lim n l a n
lowing precise deﬁnition is similar to Deﬁnition 4.4.7.
5 Definition lim n l an
integer N such that . The fol means that for every positive number M there is an
an M whenever n N If lim n l a n
, then the sequence a n is divergent but in a special way. We say that
a n diverges to . 5E12(pp 736745) 1/18/06 10:08 AM Page 741 S ECTION 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 bn lim a n bn lim ca n c lim a n nl nl nl lim a n
lim a n lim bn nl lim c nl lim a n bn nl nl nl c nl lim a n nl lim lim bn nl lim bn nl nl lim a n an
bn nl if lim bn lim bn 0 if p 0 and a n nl nl [ lim a ] lim a np nl p n nl 0 The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7). Squeeze Theorem for Sequences If a n bn cn for n n 0 and lim a n lim cn nl L, then lim bn nl nl L. 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 Theorem If lim a n 0, then lim a n nl 0. nl n FIGURE 7 The sequence bn is squeezed
between the sequences a n
and cn . EXAMPLE 4 Find lim
nl n
n 1 . 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 nl n
n 1 nl 1
1  This shows that the guess we made earlier
from Figures 1 and 2 was correct. 1
Here we used Equation 4 with r 1. lim 1 1 lim 0 nl 1
n
1 lim 1 nl lim nl 1
n 5E12(pp 736745) 742 ❙❙❙❙ 1/18/06 10:08 AM Page 742 CHAPTER 12 INFINITE SEQUENCES AND SERIES EXAMPLE 5 Calculate lim
nl ln n
.
n SOLUTION Notice that both numerator and denominator approach inﬁnity as n l . 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
fx
ln x x and obtain lim xl ln x
x lim xl 1x
1 0 Therefore, by Theorem 3 we have
ln n
n lim nl an 0
1 n is convergent or divergent. EXAMPLE 6 Determine whether the sequence a n 1 SOLUTION If we write out the terms of the sequence, we obtain 0 1 2 3 4 n 1, 1, 1, 1, 1, 1, 1, . . . _1 The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and
1 inﬁnitely often, a n does not approach any number. Thus, lim n l
1 n does not exist;
n
1 is divergent.
that is, the sequence FIGURE 8
 The graph of the sequence in Example 7 is
shown in Figure 9 and supports the answer. EXAMPLE 7 Evaluate lim
nl 1
n n if it exists. SOLUTION an
1 1
n lim nl n lim nl 1
n 0 Therefore, by Theorem 6,
0 1 n lim nl 1
n n 0 _1 n! n n, where EXAMPLE 8 Discuss the convergence of the sequence a n
FIGURE 9 n! 123 n. SOLUTION Both numerator and denominator approach inﬁnity as n l 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 7 a2 1 an 12
22 a3 123
nnn 123
333 n
n It appears from these expressions and the graph in Figure 10 that the terms are decreas 5E12(pp 736745) 1/18/06 10:08 AM Page 743 ❙❙❙❙ S ECTION 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 y ing and perhaps approach 0. To conﬁrm this, observe from Equation 7 that
1
n an 23
nn n
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 an
n t! t t and graphing in dot mode starting with t 1,
setting the tstep equal to 1. The result is shown
in Figure 10. We know that 1 n l 0 as n l . Therefore, a n l 0 as n l
Theorem. 1 by the Squeeze 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 a x
for a 1 and lim x l a x 0 for
0 a 1. Therefore, putting a r and using Theorem 2, we have
0 10 lim r n if r
if 0 0 nl FIGURE 10 1
r 1 It is obvious that
lim 1n nl If 1 r 0, then 0 lim 0 n and 1 0 nl 1, so r lim r n lim r nl n 0 nl and therefore lim n l r n 0 by Theorem 6. If r
1, then r n diverges as in
Example 6. Figure 11 shows the graphs for various values of r. (The case r
1 is
shown in Figure 8.)
an an r>1
1 _1<r<0 1 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
values of r.
8 lim r n nl 0
1 1 r 1 and divergent for all other if 1 r
if r 1 1 5E12(pp 736745) 744 ❙❙❙❙ 1/18/06 10:08 AM Page 744 CHAPTER 12 INFINITE SEQUENCES AND SERIES a n 1 for all n 1, that
9 Definition A sequence an is called increasing if a n
is, a1 a2 a3
. It is called decreasing if a n a n 1 for all n 1. It is
called monotonic if it is either increasing or decreasing. 3 EXAMPLE 10 The sequence n is decreasing because 5
3 n
for all n 5 3
1 n 3
5 n 6 1. (The right side is smaller because it has a larger denominator.)
n EXAMPLE 11 Show that the sequence a n
SOLUTION 1 We must show that a n n2 is decreasing. 1 a n , that is, 1 n
n 1
1 n 2 n2 1 1 This inequality is equivalent to the one we get by crossmultiplication:
n 1 n
1 n 1 n3
1 Since n 1, we know that the inequality n 2
so a n is decreasing. x2
x Thus, f is decreasing on 1, x 2x 2 2 n n2
n 1 1 2 nn
1 1 n3 1 2n 2 2n n
1 is true. Therefore, a n 1 a n and : 1
x2 x2
12 0 and so f n fn 1 . Therefore, a n is decreasing. 1
2 n2 x SOLUTION 2 Consider the function f x fx 1 n2 n &? 1 2 &?
&? n 2 1 2 whenever x 2 1 1 0 Definition A sequence a n is bounded above if there is a number M such that an M for all n 1 It is bounded below if there is a number m such that
m an for all n 1 If it is bounded above and below, then a n is a bounded sequence. 5E12(pp 736745) 1/18/06 10:08 AM Page 745 S ECTION 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 0 but not above. The
sequence a n n n 1 is bounded because 0 a n 1 for all n.
We know that not every bounded sequence is convergent [for instance, the sequence
an
1 n satisﬁes 1 a n 1 but is divergent from Example 6] and not every monotonic sequence is convergent a n n l . 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 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 of real
numbers, which says that if S is a nonempty set of real numbers that has an upper bound
M (x 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 M .) The Completeness Axiom is
an expression of the fact that there is no gap or hole in the real number line.
1 1 Monotonic Sequence Theorem Every bounded, monotonic sequence is convergent. Proof Suppose a n is an increasing sequence. Since a n is bounded, the set S
a n n 1 has an upper bound. By the Completeness Axiom it has a least upper
bound L . Given
is not an upper bound for S (since L is the least upper
0, L
bound). Therefore
aN L for some integer N But the sequence is increasing so a n a N for every n an since a n L N we have L 0 so N . Thus, if n an L . Thus
L an whenever n N so lim n l 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 an 1 1
2 an 6 for n 1, 2, 3, . . . SOLUTION We begin by computing the ﬁrst several terms: a2 a1 2 a4 1
2 a7 5.9375 5 6 5.5 1
2 a3 1
2 a5 5.75 a6 5.875 a8 5.96875 a9 5.984375 2 6 4 4 6 5 5E12(pp 746755) 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 n 1 a n for all n 1. This is true for n 1 because a 2 4 a 1. If we assume that it
is true for n k, then we have
ak
so ak
1
2 and 1 6 1 ak ak
ak
1
2 6 1 Thus ak 6
ak ak 2 6 1 We have deduced that a n 1 a n is true for n k 1. Therefore, the inequality is true
for all n by induction.
Next we verify that a n is bounded by showing that a n 6 for all n. (Since the
sequence is increasing, we already know that it has a lower bound: a n a 1 2 for
all n.) We know that a 1 6, so the assertion is true for n 1. Suppose it is true for
n k. Then
ak
so 6 ak
1
2 ak 6
12
1
2 6 Thus ak 12 6 6 1 This shows, by mathematical induction, that a n 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 an exists, we can use the recurrence relation to write
1
2 an  A proof of this fact is requested in
Exercise 52. Since a n l L , it follows that a n 1 l L , too (as n l , n 1 nl 1
2 6 1
2 L
Solving this equation for L, we get L  12.1 ( lim a lim lim a n nl L nl n ) 6 1
2 L 6 1 l , too). So we have 6 6, as predicted. Exercises 1. (a) What is a sequence? (b) What does it mean to say that lim n l a n
(c) What does it mean to say that lim n l a n 3–8 8?
? 2. (a) What is a convergent sequence? Give two examples. (b) What is a divergent sequence? Give two examples.  3. a n
5. a n List the ﬁrst ﬁve terms of the sequence.
1
3 0.2
1
n! n 4. a n n
3n 1
1 n 6. 2 4 6 2n 5E12(pp 746755) 1/18/06 10:52 AM Page 747 S ECTION 12.1 SEQUENCES 7. a 1
■ 3,
■ 9–14 an 2a n 1 ■ ■ 8. a 1 1 ■ ■ ■ 4, ■ an an 1 ■ an ■ ■ 47. a n  1
{1 , 1 , 1 , 16 , . . .}
248 10. 11. 2, 7, 12, 17, . . .
13. {1, ■ 24
39 ■ , . . .} 8
27 ,,
■ 12. ■ 48. a n {1 , 1 , 1 , 1 , . . . }
2468
34
{ 1 , 2 , 16 , 25 , . . .}
49 ■ ■ ■ ■ ■ ■ ■  Determine whether the sequence converges or diverges.
If it converges, ﬁnd the limit. nn 16. a n 1
5n 2
n2 17. a n 3
n 19. a n 2n
3n 1 20. a n n
1 n2 23. a n 22. a n
24. a n cos n 2
2n
2n 25. 1!
1! 29. n 2e ■ 1
2 an
3a n 1 ■ ■ ■ ■ ■ ■ if a n is an even number
if a n is an odd number 1 lim a n cos 2 n 25. Make a conjecture about 1 lim a n nl (b) A sequence a n is deﬁned by a 1 1 and
a n 1 1 1 a n for n 1. Assuming that a n is convergent, ﬁnd its limit.
53. Suppose you know that a n is a decreasing sequence and 28. n 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? ln n
ln 2 n 30. n cos n
32. a n ln n 1 54–60  Determine whether the sequence is increasing,
decreasing, or not monotonic. Is the sequence bounded? ln n 33. a n n sin 1 n 34. a n sn sn 2 1 36. a n sin 2 n
1 sn 1n 2
n 1 38. 37. 0, 1, 0, 0, 1, 0, 0, 0, 1, . . . n!
2n
■ {, ■ ■ ■ arctan 1 n n 1
n 2n
2n 1 3
n! ■ ■ ■ 42. a n 44. 2n
3n 58. a n ne 60. a n n 55. a n 3
4 57. a n 1
2n n ■ ■ 2 sin n
sn 2 n ■ ■ ■ n 59. a n 3 cos n , , , , , , , . . .}  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.) 41. a n
1
5n 11111111
13243546 40. a n
■ 54. a n
56. a n ; 41–48 43. ■ nl cos n
2n ■ ■ 52. (a) If a n is convergent, show that 31. a n 39. a n ■ 1 n n3
2n2 1 2 35. a n n! ■ 51. For what values of r is the sequence nr n convergent? 26. arctan 2 n en e n
e 2n 1 27. 1 and a1 11. Do the same if a1
this type of sequence. sn n3 1 2n 135 an n
1 2n
n 2n 50. Find the ﬁrst 40 terms of the sequence deﬁned by sn
sn 1 135 1
1 n1 1 21. a n 18. a n n
3n 5n after n years the investment is worth a n 1000 1.06 n 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 n
s3 n 49. If $1000 is invested at 6% interest, compounded annually, then 14. 5, 1, 5, 1, 5, 1, . . .
■ 46. a n 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! 45. a n 1 ❙❙❙❙ 2 n
n2 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 n 1 s2 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 a n exists.
(b) Find lim n l a n. ■ 5E12(pp 746755) 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 n 1 3 1 a n is
3 for all n. Deduce that a n is convergent increasing and a n
and ﬁnd its limit. ( f) Use Theorem 11 to show that lim n l
(The limit is e. See Equation 7.4.9.) a1 b. Let a 1 be their
arithmetic mean and b1 their geometric mean: an 2 1
1 3 a a1 an satisﬁes 0 a n 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 fn 1 fn and show that a n 1 1 1 a n 2. Assuming that a n is convergent, ﬁnd its limit.
66. (a) Let a 1 a, a 2 f a , a 3 f a 2
f f a ,...,
a n 1 f a n , where f is a continuous function. If
lim n l a n L, show that f L
L.
(b) Illustrate part (a) by taking f x
cos x, a 1, and estimating the value of L to ﬁve decimal places. lim nl sab b1 Repeat this process so that, in general,
an bn an
1 bn 2 sa n bn 1 (a) Use mathematical induction to show that
an an bn 1 bn 1 (b) Deduce that both a n and bn are convergent.
(c) Show that lim n l a n lim n l bn . Gauss called the
common value of these limits the arithmeticgeometric
mean of the numbers a and b.
72. (a) Show that if lim n l a 2 n L and lim n l a2n
a n is convergent and lim n l a n L.
(b) If a 1 1 and
an 5 b
2 ; 67. (a) Use a graph to guess the value of the limit 1 1 L, then 1 1 1 an ﬁnd the ﬁrst eight terms of the sequence a n . Then use
part (a) to show that lim n l a n s2. This gives the
continued fraction expansion n
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.
68. Use Deﬁnition 2 directly to prove that lim n l r n 1 1 s2 2 0 1
2 73. The size of an undisturbed ﬁsh population has been modeled 1. by the formula 69. Prove Theorem 6. [Hint: Use either Deﬁnition 2 or the Squeeze Theorem.]
1
n
(a) Show that if 0 70. Let a n 1 n n exists. 71. Let a and b be positive numbers with a 64. Show that the sequence deﬁned by when r 1 1 bn n .
a
1 b b, then
an
a 1 n 1 bn (b) Deduce that b n n 1 a n b
a n 1.
(c) Use a 1 1 n 1 and b 1 1 n in part (b) to
show that a n is increasing.
(d) Use a 1 and b 1 1 2 n in part (b) to show that
a 2 n 4.
(e) Use parts (c) and (d) to show that a n 4 for all n. pn 1 a b pn
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 0.
(a) Show that if pn is convergent, then the only possible
values for its limit are 0 and b a.
(b) Show that pn 1
b a pn.
(c) Use part (b) to show that if a b, then lim n l pn 0; in
other words, the population dies out.
(d) Now assume that a b. Show that if p 0 b a, then
pn is increasing and 0 pn b a. Show also that if
p 0 b a, then pn is decreasing and pn b a.
Deduce that if a b, then lim n l pn b a. 5E12(pp 746755) 1/18/06 10:52 AM Page 749 S ECTION 12.2 SERIES ❙❙❙❙ 749 LABORATORY PROJECT
C AS L ogistic Sequences
A sequence that arises in ecology as a model for population growth is deﬁned by the logistic
difference equation
pn 1 k pn 1 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 pn 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 p0 1. Use this program to do the following.
1. Calculate 20 or 30 terms of the sequence for p0 1
2 and for two values of k such that
1 k 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
a1 1 a2 a3 n1 we get an expression of the form an which is called an inﬁnite series (or just a series) and is denoted, for short, by the symbol
an or an n1 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 2 3 4 5 n 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 n n 1 2, which becomes very large as n increases.
However, if we start to add the terms of the series
1
2 1
4 1
8 1
16 1
32 1
64 1
2n 5E12(pp 746755) ❙❙❙❙ 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 1 2 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 n1 1
2n 1
2 1
4 1
8 1
16 1
2n 1 We use a similar idea to determine whether or not a general series (1) has a sum. We
consider the partial sums
s1 a1 s2 a1 a2 s3 a1 a2 a3 s4 a1 a2 a3 a4 and, in general,
n sn a1 a2 a3 an ai
i1 These partial sums form a new sequence sn , which may or may not have a limit. If
lim n l 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 n1 an a1 a2 , let sn denote its nth a3 partial sum:
n sn ai a1 a2 an i1 If the sequence sn is convergent and lim n l sn s exists as a real number, then
the series a n is called convergent and we write
a1 a2 an s or an s n1 The number s is called the sum of the series. Otherwise, the series is called divergent.
Thus, when we write n 1 a n s we mean that by adding sufﬁciently many terms of
the series we can get as close as we like to the number s. Notice that
 Compare with the improper integral
n y 1 f x dx lim tl y t 1 an f x dx To ﬁnd this integral we integrate from 1 to t and
then let t l . For a series, we sum from 1 to n
and then let n l . n1 lim nl ai
i1 EXAMPLE 1 An important example of an inﬁnite series is the geometric series a ar ar 2 ar 3 ar n 1 ar n
n1 1 a 0 5E12(pp 746755) 1/18/06 10:52 AM Page 751 S ECTION 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 ar so s a
1 r ar a rsn ar 2 ar n 1 ar sn ar 2 ar n 1 and ar@ ar n Subtracting these equations, we get ar@ sn 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 a
a na l
. Since lim n l sn doesn’t exist, the
geometric series diverges in this case.
If r 1, we have ar# aar ❙❙❙❙ s a If 1 ar n a rn
r a1
1 sn 3 1, we know from (12.1.8) that r n l 0 as n l , so r lim sn nl a rsn lim nl rn
r a1
1 a
1 a
r 1 r lim r n nl a
1 r Thus, when r
1 the geometric series is convergent and its sum is a 1 r .
If r
1 or r 1, the sequence r n is divergent by (12.1.8) and so, by Equation 3,
lim n l sn does not exist. Therefore, the geometric series diverges in those cases. FIGURE 1 We summarize the results of Example 1 as follows. 4 The geometric series
ar n 1 a ar 2 ar n1  In words: The sum of a convergent geometric
series is
1 is convergent if r 1 and its sum is first term
common ratio ar n a 1 1 n1 If r r r 1 1, the geometric series is divergent. EXAMPLE 2 Find the sum of the geometric series 5 10
3 20
9 40
27 SOLUTION The ﬁrst term is a 5 and the common ratio is r
series is convergent by (4) and its sum is
5 10
3 20
9 40
27 5
1 ( 2)
3 2
3 . Since r 5
5
3 3 2
3 1, the 5E12(pp 746755) 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 22n31 EXAMPLE 3 Is the series n convergent or divergent? n1 SOLUTION Let’s rewrite the nth term of the series in the form ar n 1:
 Another way to identify a and r is to write
out the ﬁrst few terms:
4 16
3 22n31 64
9 4n
3n 1 n n1 n1 n1 4( 4 )
3
n1 We recognize this series as a geometric series with a
series diverges by (4).
EXAMPLE 4 Write the number 2.317 4 and r 4
3 . Since r 1, the 2.3171717. . . as a ratio of integers. SOLUTION 2.3171717. . . 17
10 3 2.3 17
10 5 17
10 7 After the ﬁrst term we have a geometric series with a 17 10 3 and r
Therefore
17
17
3
10
1000
2.317 2.3
2.3
1
99
1
2
10
100
23
17
1147
10
990
495
x n, where x EXAMPLE 5 Find the sum of the series 1 10 2. 1. n0 0 and so the ﬁrst term is x 0
1 even when x 0.) Thus SOLUTION Notice that this series starts with n
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
xn 1 0 x2 x x3 1. (With x4 n0 This is a geometric series with a
(4) gives 1 and r xn 5
n0 x. Since r
1
1 x x 1, it converges and 5E12(pp 746755) 1/18/06 10:53 AM Page 753 S ECTION 12.2 SERIES 1 EXAMPLE 6 Show that the series nn n1 ❙❙❙❙ 753 is convergent, and ﬁnd its sum. 1 SOLUTION This is not a geometric series, so we go back to the deﬁnition of a convergent
series and compute the partial sums.
n 1 sn ii i1 1 1 12 1 1 1 23 34 nn 1 We can simplify this expression if we use the partial fraction decomposition
1
ii 1
i 1 1
i 1 (see Section 8.4). Thus, we have
n
i1 ii 1 1 1
2 n i 1 1
3 1
3 1
4 1
n 1
n 1
lim sn lim nl 1 1 nl n 1 1 0 ) 1 1 Therefore, the given series is convergent and
1
nn n1 1 1 EXAMPLE 7 Show that the harmonic series 1
n1 sn 1 1 and so
 Figure 3 illustrates Example 6 by showing
the graphs of the sequence of terms
a n 1 [n n 1 ] 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
i i1 1
2 1  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
oldfashioned collapsing telescope) into just two
terms. n 1 sn 1
n 1
2 1 1
3 1
4 is divergent.
SOLUTION s1
s2 an
0 FIGURE 3 n 1
1 1
2 s4 1 1
2 (1
3 1
4 s8 1 1
2 1
4 1 1
2 (1
3
(1
4 1 1
2 1
2 1 1
2 1 1
2 (1
3
(1
4 1 1
2 1
2 s16 ) 1 )
1
4) (1
5
(1
8 1
2 1
6 1
7 1
8 1
8 1
8 1
8 (1
5
(1
8 1
4 1
2 1
4 1
2 2
2 )
) 3
2 1 )
1
4) (1
4 1
2 )
1
8)
1
8 1 4
2 (1
9
1
( 16 1
16 )
) 1
16 1 5E12(pp 746755) 754 ❙❙❙❙ 1/18/06 10:53 AM Page 754 CHAPTER 12 INFINITE SEQUENCES AND SERIES Similarly, s32 1 5
2 , s64 6
2 1 , and in general
s2 n  The method used in Example 7 for showing
that the harmonic series diverges is due to the
French scholar Nicole Oresme (1323–1382). This shows that s2 n l
series diverges. 6 as n l Theorem If the series n
2 1 and so sn is divergent. Therefore, the harmonic a n is convergent, then lim an
nl n1 0. Proof Let sn a1 a2
a n . Then a n sn sn 1. Since a n is convergent, the
sequence sn is convergent. Let lim n l sn s. Since n 1 l as n l , we also
have lim n l sn 1 s. Therefore
lim a n lim sn nl s
NOTE 1 s lim sn 1 nl lim sn nl 1 0 ■ NOTE 2 sn nl ■ 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.  The converse of Theorem 6 is not true in general. If lim n l a n 0, we cannot
conclude that a n is convergent. Observe that for the harmonic series 1 n we have
a n 1 n l 0 as n l , but we showed in Example 7 that 1 n is divergent. 7 The Test for Divergence If lim a n does not exist or if lim a n
nl series 0, then the nl a n is divergent.
n1 The Test for Divergence follows from Theorem 6 because, if the series is not divergent,
then it is convergent, and so lim n l a n 0.
n2 EXAMPLE 8 Show that the series
n1 5n 2 4 diverges. SOLUTION lim a n nl lim nl n2
5n 2 4 lim nl 5 1
4 n2 1
5 0 So the series diverges by the Test for Divergence.
NOTE 3 If we ﬁnd that lim n l a n
0, we know that a n is divergent. If we ﬁnd that
lim n l a n 0, we know nothing about the convergence or divergence of a n. Remember the warning in Note 2: If lim n l a n 0, the series a n might converge or it might
diverge.
■ 5E12(pp 746755) 1/18/06 10:53 AM Page 755 S ECTION 12.2 SERIES 8 Theorem If a n and
(where c is a constant), (i) ca n c an n1 (iii) bn are convergent series, then so are the series
a n bn , and a n bn , and
(ii) an n1 an bn an n1 bn an n1 ❙❙❙❙ 755 ca n
bn n1 n1 bn n1 n1 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
n sn n ai s an i1 tn bi n1 The nth partial sum for the series t bn i1 an n1 bn is
n un ai bi i1 and, using Equation 5.2.9, we have
n lim u n nl n lim nl ai bi lim nl i1
n ai lim sn an lim bi
i1 bi nl i1 i1 lim tn nl Therefore, i1 n lim nl n ai s nl t bn is convergent and its sum is
an bn s t an n1 n1 3 EXAMPLE 9 Find the sum of the series
n1 SOLUTION The series bn n1 nn 1
.
2n 1 1
2 n 1 2 is a geometric series with a n1 1
2 1
2n 1
2 , so 1 1
2 1 and r In Example 6 we found that
1
n1 nn 1 1 So, by Theorem 8, the given series is convergent and
3
n1 nn 1 1
2n 1 3
n1 31 nn 1 1 4 n1 1
2n 5E12(pp 756765) 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
■ n
n3 1 2
9 n4 3
28 is convergent. Since
n
n1 n 3 1
2 1 n
n4 n 3 1 it follows that the entire series n 1 n n 3 1 is convergent. Similarly, if it is known that
the series n N 1 a n converges, then the full series
N an an n1 n1 an
nN1 is also convergent.  12.2 Exercises
14. 1 1. (a) What is the difference between a sequence and a series? 0.4 0.16 0.064 (b) What is a convergent series? What is a divergent series?
2. Explain what it means to say that n 5( 2 )
3 15. 5. 1 an  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. n1 3n
4n 17.
n1 n1 5. 4. n n1 6. tan n
n1 1
1 n1 1
n 1.5 ■ 1 ■ n 1.5 1 ■ ■ ■ nn ■ 1
■ ■ ■ 2n
.
3n 1
(a) Determine whether a n is convergent.
(b) Determine whether n 1 a n is convergent. ■ 13. 4
3 2 5
2 8
9
25
8 n 0.3 n ln 32. arctan n 2n cos 1 5
k k1 3
nn ■ ■ ■ n1 n1 30. n1 aj 0.8 3 5
4n 3
■ ■ 34.
n1
■ ■ 3
5n ■ 2
n
■ ■ ■ ■ i1 12.
125
32 2
4n n1 33.
n Determine whether the series is convergent or divergent.
If it is convergent, ﬁnd its sum.
2 n2 28. n n1  11. 3 12
2 n1 2n
6 31. aj and n
nn 26. n
s2 35–40
11–34 n1 1 3n 29. (b) Explain the difference between
ai 3
n 24. 1
k n1 j1 i1 n1 2 k2 k2 27. n n en
3n 1 22. 5 n1 and i1 n1 n2 25. 10. (a) Explain the difference between
n (s2 ) n 20. 2
n2 9. Let a n ai n0 n 23. 1
n2 ■ 1 18. n1 n 1 n1 0.6 8. 3 21. n1 7.
■ 2n2
n2 n0 n1 12
5 1 n 19. n1 6
5n 16. n1 ; 3–8 3. n1 1
8 1
4 1
2 1  35. 0.2 Express the number as a ratio of integers.
36. 0.73 0.2222 . . . 37. 3.417 38. 6.254 3.417417417 . . . 39. 0.123456
■ ■ ■ 0.73737373 . . . 40. 5.6021
■ ■ ■ ■ ■ ■ ■ ■ ■ 5E12(pp 756765) 1/18/06 10:09 AM Page 757 ❙❙❙❙ S ECTION 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. n1 43. xn
3n 42. 4nx n 41. 44. 4 (a) Assuming that the ball continues to bounce indeﬁnitely,
ﬁnd the total distance that it travels. (Use the fact that the
ball falls 1 t t 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 k v, where
0 k 1. How long will it take for the ball to come
to rest? n 3 x n n1 n0 x
2 n0 n n 45.
n0
■ cos x
2n ■ 53. What is the value of c if ■ ■ ■ ■ ■ ■ ■ ■ ■ 1
n 1
n1 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. 1 47.
n1
■ ■ 4n 1 4n
■ 3 ■ n1 ■ ■ ■ 49. If the nth partial sum of a series sn
ﬁnd a n and n1 n
n n1 3n
n2 ■ n1 1
n ■ 2
■ ■ ■ nn 1 1 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 1
1 an . 50. If the nth partial sum of a series ﬁnd a n and n2 48. 2? x n, 0 x 1, for n 0, 1, 2, 3, 4, . . .
on a common screen. By ﬁnding the areas between successive
curves, give a geometric demonstration of the fact, shown in
Example 6, that is another series with this property.
C AS n c ; 54. Graph the curves y whose terms approach 0. Show that
ln 1 1
n2 ■ 46. We have seen that the harmonic series is a divergent series n1 757 n1 an is sn 3 P n 2 n, 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 twicespent 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 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 Sn kD, where k 1 s. 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 56. A right triangle ABC is given with A
and AC
b.
CD is drawn perpendicular to AB, DE is drawn perpendicular
to BC, EF AB, and this process is continued indeﬁnitely as
shown in the ﬁgure. Find the total length of all the
perpendiculars
CD DE EF FG 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 r 1. Suppose that the ball is dropped from an initial height of H meters. B G E C 5E12(pp 756765) 758 ❙❙❙❙ 1/18/06 10:10 AM Page 758 CHAPTER 12 INFINITE SEQUENCES AND SERIES nitely many numbers. Give examples of some numbers in
the Cantor set.
(b) The Sierpinski carpet is a twodimensional 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. 57. What is wrong with the following calculation? 0 0 0 1 0 1 1 1 1 1 1
1 1 0 1
1 1
1 1
0 1 1
1 1 0 1 1 1 (Guido Ubaldus thought that this proved the existence of God
because “something has been created out of nothing.”)
58. Suppose that n1 series. Prove that a n a n 0 is known to be a convergent
n 1 1 a n is a divergent series. 59. Prove part (i) of Theorem 8.
60. If
61. If a n is divergent and c 0, show that ca n is divergent. a n is convergent and bn is divergent, show that the series
a n bn is divergent. [Hint: Argue by contradiction.] 62. If a n and
divergent? bn are both divergent, is an bn necessarily 63. Suppose that a series a n has positive terms and its partial
sums sn satisfy the inequality sn 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 fn 1 fn 2 n 3 Show that each of the following statements is true.
1
1
1
(a)
fn 1 fn 1
fn 1 fn
fn fn 1 n2 1
fn 1 fn n2 fn (b)
(c) fn
f 1
1 2 1n1 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 ).
33
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 n 1 a n 2 for n 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 a n in terms of a 1 and a 2 by expressing
a n 1 a n in terms of a 2 a 1 and summing a series.
67. Consider the series n
n1 n 1! (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. 5E12(pp 756765) 1/18/06 10:10 AM Page 759 S ECTION 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 1 n n 1 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 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
i1 5
10
50
100
500
1000
5000 1
i2 1.4636
1.5498
1.6251
1.6350
1.6429
1.6439
1.6447 n1 1
n2 1
12 1
22 1
32 1
42 1
52 There’s no simple formula for the sum sn of the ﬁrst n terms, but the computergenerated
table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as n l 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 1 x 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 1 x 2 at the right endpoint
of the interval. So the sum of the areas of the rectangles is
1
12 1
22 y 1
32 1
42 1
52 n1 1
n2 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 1 x 2 for x 1, which is the value of the integral
x1 1 x 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
1
12 y 1 1
dx
x2 2 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: n1 1
n2 1
12 1
22 1
32 1
42 2 5E12(pp 756765) 760 ❙❙❙❙ 1/18/06 10:10 AM Page 760 CHAPTER 12 INFINITE SEQUENCES AND SERIES n n sn
i1 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 2 6, 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 n1 1
sn 1
s1 1
s2 1
s3 1
s4 1
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 1 sx, 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
œ2
„ 4 area= 1
œ3
„ 5 x area= 1
œ4
„ The base of each rectangle is an interval of length 1. The height is equal to the value of
the function y 1 sx at the left endpoint of the interval. So the sum of the areas of all the
rectangles is
1
1
1
1
1
1
s1
s2
s3
s4
s5
n 1 sn
This total area is greater than the area under the curve y 1 sx for x 1, which is equal
to the integral x1 (1 sx ) 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,
and let a n f n . Then the series n 1 a n is convergent if and only if the improper
integral x1 f x d x is convergent. In other words: (i) If y f x d x is convergent, then
1 a n is convergent.
n1 (ii) If y f x d x is divergent, then
1 NOTE ■ gral at n a n is divergent.
n1 When we use the Integral Test it is not necessary to start the series or the inte1. For instance, in testing the series
1
n4 n 3 2 we use y 4 1
x 3 2 dx 5E12(pp 756765) 1/18/06 10:10 AM Page 761 S ECTION 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 n N a n is
convergent, so n 1 a n is convergent by Note 4 of Section 12.2.
1 EXAMPLE 1 Test the series
n1 n2 1 1 x2
so we use the Integral Test: SOLUTION The function f x 1, 1 y x2 1 1 dx for convergence or divergence.
1 is continuous, positive, and decreasing on lim y tl 1 t 1 x2 1 lim tan 1t tl 4 tl lim tan 1x dx 2 t 1 4 4 Thus, x1 1 x 2 1 d x is a convergent integral and so, by the Integral Test, the series
1 n 2 1 is convergent.
1
convergent?
np EXAMPLE 2 For what values of p is the series
n1
p 0, then lim n l 1 n
. If p 0, then lim n l 1 n p
1. In either
p
case lim n l 1 n
0, so the given series diverges by the Test for Divergence (12.2.7).
If p 0, then the function f x
1 x p is clearly continuous, positive, and decreasing
on 1, . We found in Chapter 8 [see (8.8.2)] that
SOLUTION If p y 1 1
dx converges if p
xp 1 and diverges if p 1 It follows from the Integral Test that the series 1 n p converges if p 1 and diverges
if 0 p 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 pseries. It is important in the rest of this chapter,
so we summarize the results of Example 2 for future reference as follows. 1 The pseries
n1 1
is convergent if p
np 1 and divergent if p EXAMPLE 3 (a) The series n1 1
n3 1
13 1
23 1
33 is convergent because it is a pseries with p
(b) The series n1 1
n1 3 n1 1
3
sn 1 is divergent because it is a pseries with p 3 1. 1
3
s2
1
3 1
43 1
3
s3
1. 1
3
s4 1. 5E12(pp 756765) 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,
■ n1 2 1
n2 6 1
dx
x2 y whereas 1 1 Therefore, in general, y an 1 n1 f x dx ln n
converges or diverges.
n EXAMPLE 4 Determine whether the series
n1 SOLUTION The function f x
ln x x is positive and continuous for x 1 because the
logarithm function is continuous. But it is not obvious whether or not f is decreasing, so
we compute its derivative: 1 x x ln x
x2 fx Thus, f x
0 when ln x 1, that is, x
and so we can apply the Integral Test: y 1 ln x
dx
x x ln x
dx
x t 1 ln t
lim
tl
2 tl 0 x y=ƒ Rn Rn FIGURE 4 ln n n is also divergent by the s sn an 1 an an 2 3 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, . Comparing the areas of the rectangles with the area under y f x for x n
in Figure 3, we see that y n+1 1 2 ... FIGURE 3 0 t 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 sn s. But how good is such an
approximation? To ﬁnd out, we need to estimate the size of the remainder n an+1 an+2 2 e Estimating the Sum of a Series y=ƒ an+1 an+2 ln x
2 lim Since this improper integral is divergent, the series
Integral Test.
y ln x
2 e. It follows that f is decreasing when x lim y tl 1 ... an 1 an 2 y f x dx y f x dx n Similarly, we see from Figure 4 that
x Rn an 1 an 2 n1 5E12(pp 756765) 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 k ous, positive, decreasing function for x
then y f x dx n1 a k, where f is a continua n is convergent. If Rn s sn , n and y Rn f x dx n EXAMPLE 5 (a) Approximate the sum of the series 1 n 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 xn f x d x. With f x 1 x 3, we have y n 1
dx
x3 t 1
2x 2 lim tl 1
2t 2 lim tl n 1
2n 2 1
2n 2 (a) n1 1
n3 1
13 s10 1
23 1
33 1
10 3 1.1975 According to the remainder estimate in (2), we have y R10 10 1
dx
x3 1
2 10 1
200 2 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 0.0005. Since
Rn y n 1
2n 2 we want 1
dx
x3 1
2n 2 0.0005 Solving this inequality, we get
n2 1
0.001 1000 or n s1000 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 y n1 f x dx s sn y n f x dx 31.6 5E12(pp 756765) 764 ❙❙❙❙ 1/18/06 10:10 AM Page 764 CHAPTER 12 INFINITE SEQUENCES AND SERIES because sn 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.
EXAMPLE 6 Use (3) with n 10 to estimate the sum of the series
n1 1
.
n3 SOLUTION The inequalities in (3) become y s10 11 1
dx
x3 s y s10 10 1
dx
x3 From Example 5 we know that y n so 1
2 11 s10 1
dx
x3 1
2n 2
1
2 10 s s10 1.201664 Using s10 2 s 1.202532 2 1.197532, we get If we approximate s by the midpoint of this interval, then the error is at most half the
length of the interval. So n1 y 1
n3 1.2021 with error 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 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 ... nx 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 1 n 2 and 1 sn. 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, 2 ,
that is, f 2
a 2 . So, comparing the areas of the shaded rectangles with the area under
y f x from 1 to n, we see that y=ƒ
4
an1
a¡ a™ a£ a¢ 0 1 FIGURE 6 2 3 4 5 ... a2 a3 an y n 1 f x dx (Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6
shows that nx 5 y n 1 f x dx a1 a2 an 1 5E12(pp 756765) 1/18/06 10:10 AM Page 765 S ECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 765 (i) If y f x d x is convergent, then (4) gives
1 n y ai y f x dx 1 i2 since f x n 1 f x dx 0. Therefore
n sn a1 ai y a1 Since sn f x dx 1 i2 M, say M for all n, the sequence sn is bounded above. Also
sn sn 1 an 1 sn since a n 1 f n 1
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 x1 f x d x is divergent, then x1n f x d x l as n l because f x
0. But (5)
gives y n 1 and so sn  12.3 1
n 1.3 1 1
8 12. 1 1
dx
x 1.3 y 1
2 s2 6 1 5 n1 6 f x dx ai ai i1 n1 1
n4 4.
n1 n 7. n1  9.
n1 n1 n ■ ■ ■ ■ 3n n1 ne n
n 8.
■ ■ ■ n2
n 10. n
n1 1.4 3n 1
4 s4 1
5 s5 n3 n1 ■ 25.
n2 2
1
1
4n 2 n1 ln n
n2 n1 n4 n3 1
n ln n ln ln n 24. n
■ ■ ■ ■ 5 n 22. ■  n 20. 1 25–28 3n
nn 18. 2 n1 n3 ■ n 16. n2 n1
■ 5 14. 1
n ln n 23. 1.2 1
3 s3 1 n2 ■ 1
125 4 2 ne 21. 2
1
■ 1
64 n1 1 Determine whether the series is convergent or divergent.
2
n 0.85 19. 1 5. n1 ■ 9–24 1
4
sn 1
27 n 17. i2 Use the Integral Test to determine whether the series is
convergent or divergent. ■ a n diverges. 1 15.  e 1 2 sn
n3 n1 x 1 and an f n . By drawing a picture, rank the following
three quantities in increasing order: y 5 13. 2. Suppose f is a continuous positive decreasing function for 6. and so 11. 1 What can you conclude about the series? n1 sn Exercises n2 3. ai
i1 l . This implies that sn l 1 1. Draw a picture to show that 3–8 n1 f x dx ■ 1 ■ ■ ■ Find the values of p for which the series is convergent. 1
n ln n p 26.
n3 1
n ln n ln ln n p ■ 5E12(pp 766775) 766 ❙❙❙❙ 27. 1/18/06 10:54 AM Page 766 CHAPTER 12 INFINITE SEQUENCES AND SERIES n2 n1 p 28. n1
■ ■ n1
■ ■ ■ ■ 29. The Riemann zetafunction ■ ln n
np ■ CAS ■ ■ ■ ■ is deﬁned by x
n1 1
nx and is used in number theory to study the distribution of prime
numbers. What is the domain of ?
1 n 4. Estimate the
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. 30. (a) Find the partial sum s10 of the series n1 ln n 2 n 2 is convergent.
(b) Find an upper bound for the error in the approximation
s 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. 36. (a) Show that the series 37. (a) Use (4) to show that if sn is the nth partial sum of the har monic series, then
sn n1 1 n 5 correct to three decimal 1
2 1 tn
n1 n 32 to within 0.01. 34. How many terms of the series n 2 1 n ln n
to add to ﬁnd its sum to within 0.01? 2 would you need 35. Show that if we want to approximate the sum of the series n 1.001 so that the error is less than 5 in the ninth decimal
place, then we need to add more than 10 11,301 terms!
n1  12.4 1
3 1
n ln n 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 x
1 x and interpret
tn as an area [or use (5)] to show that tn 0 for all n.
(b) Interpret places.
33. Estimate ln n 38. Use the following steps to show that the sequence 31. (a) Use the sum of the ﬁrst 10 terms to estimate the sum of the 32. Find the sum of the series 1 (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. tn
series n 1 1 n 2. How good is this estimate?
(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 sn is less than 0.001. n1 tn ln n 1 1 ln n 1
n 1 as a difference of areas to show that tn tn 1 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 n1 b ln n 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
1 1
n1 2 n 1 reminds us of the series n 1 1 2 n, which is a geometric series with a 1 and r 1 and
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
2n 1 1
2n 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 5E12(pp 766775) 1/18/06 10:54 AM Page 767 S ECTION 12.4 THE COMPARISON TESTS ❙❙❙❙ 767 also follows that the sum of the series is less than the sum of the geometric series:
1
2 n1 n 1 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.
The Comparison Test Suppose that (i) If
(ii) If
 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. bn is convergent and a n
bn is divergent and a n a n and bn are series with positive terms.
bn for all n, then a n is also convergent.
bn for all n, then a n is also divergent. Proof (i) Let
n n sn ai tn i1 bi t bn i1 n1 Since both series have positive terms, the sequences sn and tn are increasing
sn 1 sn a n 1 sn . Also tn l t, so tn t for all n. Since a i bi , we have sn tn .
Thus, sn 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 (since tn is increasing). But a i bi so sn tn .
Thus, sn l . Therefore, a n diverges.
Standard Series for Use
with the Comparison Test 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 pseries [ 1 n p converges if
p 1 and diverges if p 1; see (12.3.1)] or a geometric series [ ar n 1 converges if
r
1 and diverges if r
1; see (12.2.4)].
EXAMPLE 1 Determine whether the series
n1 5
4n 2n 2 3 converges or diverges. SOLUTION For large n the dominant term in the denominator is 2 n 2 so we compare the given series with the series 5 2 n 2 . Observe that
2n 2 5
4n 5
2n 2 3 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 n1 5
2n 2 5
2 n1 1
n2 is convergent because it’s a constant times a pseries with p n1 2n 2 5
4n is convergent by part (i) of the Comparison Test. 3 2 1. Therefore 5E12(pp 766775) 768 ❙❙❙❙ 1/18/06 10:54 AM Page 768 CHAPTER 12 INFINITE SEQUENCES AND SERIES bn or a n bn in the Comparison Test is given for
N OTE 1 Although the condition a n
all n, we need verify only that it holds for n 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.
■ EXAMPLE 2 Test the series
n1 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 1 for n 3 and so ln n
n 1
n n 3 We know that 1 n is divergent ( pseries with p
gent by the Comparison Test. 1). Thus, the given series is diver 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
■ 1
n1 2 n 1 The inequality
1
2 n 1 1
2n
n ( 1 ) is convergent
is useless as far as the Comparison Test is concerned because bn
2
n
1 ought to be convergent
and a n bn. Nonetheless, we have the feeling that 1 2
1n
because it is very similar to the convergent geometric series ( 2 ) . In such cases the following test can be used.
The Limit Comparison Test Suppose that
 Exercises 40 and 41 deal with the cases
c 0 and c
. a n and
lim nl where c is a ﬁnite number and c
diverge. an
bn bn are series with positive terms. If
c 0, then either both series converge or both Proof Let m and M be positive numbers such that m
c for large n, there is an integer N such that c M . Because a n bn is close to m
and so an
bn M when n N m bn an Mbn when n N 5E12(pp 766775) 1/18/06 10:54 AM Page 769 S ECTION 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.
1 EXAMPLE 3 Test the series
n1 2n for convergence or divergence. 1 SOLUTION We use the Limit Comparison Test with 1 an 2n 1
2n bn 1 and obtain
lim nl an
bn lim nl 1 2n 1
1 2n lim nl 2n
2 n lim 1 nl 1 1
1 2n 1 0 Since this limit exists and 1 2 n is a convergent geometric series, the given series converges by the Limit Comparison Test.
2n 2
s5 EXAMPLE 4 Determine whether the series
n1 3n
converges or diverges.
n5 SOLUTION The dominant part of the numerator is 2 n 2 and the dominant part of the denom inator is sn5 n 5 2. This suggests taking
2n 2
s5 an bn nl an
bn lim nl 2n 2
s5 n1 2
2 3n
n5 2 lim 2n 2
n5 2 lim 3n
n5 3
n lim nl 2 5
n5 nl 2
n1 2
2n 5 2 3n 3 2
2 s5 n 5 20
2 s0 1 1 Since bn 2 1 n 1 2 is divergent ( pseries with p
by the Limit Comparison Test. 1
2 1 1), the given series diverges Notice that in testing many series we ﬁnd a suitable comparison series
only the highest powers in the numerator and denominator. bn by keeping 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
For the comparison series s sn an 1 an 2 bn we consider the corresponding remainder
Tn t tn bn 1 bn 2 5E12(pp 766775) ❙❙❙❙ 770 1/18/06 10:54 AM Page 770 CHAPTER 12 INFINITE SEQUENCES AND SERIES Since a n bn for all n, we have Rn Tn . If bn is a pseries, 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 1 n3 1 . Estimate the error involved in this approximation. SOLUTION Since 1
n 3 1
n3 1 the given series is convergent by the Comparison Test. The remainder Tn for the comparison series 1 n 3 was estimated in Example 5 in Section 12.3 using the Remainder Estimate for the Integral Test. There we found that y Tn n 1
dx
x3 1
2n 2 Therefore, the remainder Rn for the given series satisﬁes
Rn
With n 1
2n 2 Tn 100 we have
1
2 100 R100 2 0.00005 Using a programmable calculator or a computer, we ﬁnd that
100 1
n1 n 3 1 n1 1
n3 1 0.6864538 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 bn for all n, what can you say about a n? Why?
(b) If a n bn for all n, what can you say about a n? Why? n 7. 1
n n1 a n and bn are series with positive terms and bn is
known to be divergent.
(a) If a n bn for all n, what can you say about a n ? Why?
(b) If a n bn for all n, what can you say about a n ? Why? cos n
n2 1 10.
12. n2 n2
n3 n1 n1
n4n 11.
13. 3–32  3. Determine whether the series converges or diverges. n1 n n1 2 1
n 2 5 5. 3n 1 2 4.
n1 n n2 n 3 4
1 6. 2 15. sn n1 17.
n1 n1 1
1 1
1 n
3n 4
1 sin n
10 n n0 sn 14.
n2 1
n sn 1
sn 2 n 2 n1 2. Suppose 2 n1 2 9. 3n 4 8. 2 1 n n1 1
sn 3 n1 2n n 16. 1
1 1 18. 3 5E12(pp 766775) 1/18/06 10:54 AM Page 771 S ECTION 12.5 ALTERNATING SERIES 2n 19.
n1 1 3 20. n n1 1 21.
n1 1 sn n1 5
1 n1 1
s1 23. 25. 27. n n2
n2 n6 31. 26. 38. For what values of p does the series
39. Prove that if a n 5n n1 n n
3
sn 7 2 n 2 7n
3 n 2 5n 1
n!
nn 28. 1
n! 30. 1
n n1
■ 32.
n1 ■ ■ ■ ■ ■ n lim n n1 e 1 nl 5
n2 lim 1
n1 1 n
■ ■  Use the sum of the ﬁrst 10 terms to approximate the sum
of the series. Estimate the error. 33.
n1 34. n2 n1 1 35.
n1
■ n4 ■ 1 2 n1
■ ■ ■ 42. Give an example of a pair of series a n and bn with positive
terms where lim n l a n bn
0 and bn diverges, but a n
converges. [Compare with Exercise 40.] cos n
n5 43. Show that if a n n 36. n ■ 1 ■ ■ n 13
■ an
bn then an is also divergent.
(b) Use part (a) to show that the series diverges.
1
ln n
(ii)
(i)
n
n 2 ln n
n1 33–36 1 an and bn are series with positive terms
bn is divergent. Prove that if
nl ■ 0 41. (a) Suppose that and ■ an
bn then an is also convergent.
(b) Use part (a) to show that the series converges.
ln n
ln n
(ii)
(i)
n3
sn e n
n1
n1 n ■ 2
a n also a n converges, then an and bn are series with positive terms
bn is convergent. Prove that if and 3 1 n p ln n converge? 40. (a) Suppose that n1 2 0 and n2 771 converges. 2
13 n1 1
n sin ■ n
n
n2 24. n1 n1 n3 2n
n2 2 1 29. 22. 2n
3n 1
1 ❙❙❙❙ 0 and lim n l n a n 0, then a n is divergent. n
■ ■ 44. Show that if a n ■ 0 and a n is convergent, then ln 1 an is convergent. 37. The meaning of the decimal representation of a number 45. If 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 . . .
2
3
10
10
10
10 4 a n is a convergent series with positive terms, is it true that
sin a n is also convergent? 46. 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:
1
1
2 1
2 1
3 1
4 1
5 1
6 n1 2
3 3
4 4
5 5
6 6
7 n1 1n
n
1 1 n n n 1 5E12(pp 766775) 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
an 1 n1 bn or 1 nbn an 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 1 n1 bn b1 b2 b3 b4 b5 b6 bn 0 n1 satisﬁes
(i) bn 1 bn
(ii) lim bn 0 for all n nl 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 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 s™ 0 s¢ sß s∞ s s£ s¡ Proof of the Alternating Series Test We ﬁrst consider the even partial sums: s2 s2 n b2 0 s4
In general b1
s2 b3 b4 s2 1 b2 n s2 n s2 n 2 0 Thus b2 n
s2 s4 since b2 s6 since b4
2 b1
b3
b2 n since b2 n 1 s2 n But we can also write
s2 n b1 b2 b3 b4 b5 b2 n 2 b2 n 1 b2 n 5E12(pp 766775) 1/18/06 10:55 AM Page 773 S ECTION 12.5 ALTERNATING SERIES ❙❙❙❙ 773 Every term in brackets is positive, so s2 n b1 for all n. Therefore, the sequence s2 n 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 s2 n s nl Now we compute the limit of the odd partial sums:
lim s2 n nl lim s2 n 1 b2 n nl lim s2 n lim b2 n nl s 1 nl 1 0 [by condition (ii)] s
Since both the even and odd partial sums converge to s, we have lim n l sn
Exercise 72 in Section 12.1) and so the series is convergent.
 Figure 2 illustrates Example 1 by showing
1 n 1 n and
the graphs of the terms a n
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 0.693 (see Exercise 36). 1
2 1 an 1 1
4 bn 1n
n n1 satisﬁes EXAMPLE 2 The series 1 because n1 1 1
n 1 n 3n
is alternating but
4n 1 n lim bn nl FIGURE 2 1
3 n1
1
(ii) lim bn
lim
0
nl
nl
n
so the series is convergent by the Alternating Series Test. sn 0 EXAMPLE 1 The alternating harmonic series (i) bn 1 s (see lim nl 3n
4n 1 3 lim nl 4 1
n 3
4 so condition (ii) is not satisﬁed. Instead, we look at the limit of the n th term of the series:
lim a n lim nl nl 1 n 3n
4n 1 This limit does not exist, so the series diverges by the Test for Divergence.
EXAMPLE 3 Test the series 1
n1 n2 n1 n3 1 for convergence or divergence. 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 2 n 3 1 is decreasing. However, if we consider the related function 5E12(pp 766775) 774 ❙❙❙❙ 1/18/06 10:55 AM Page 774 CHAPTER 12 INFINITE SEQUENCES AND SERIES x2 x3 fx 1 , we ﬁnd that  Instead of verifying condition (i) of the Alternating Series Test by computing a derivative, we
could verify that bn 1 bn directly by using the
technique of Solution 1 of Example 11 in
Section 12.1. x3
12 x2
x3 fx Since we are considering only positive x, we see that f x
0 if 2 x 3 0, that is,
3
3
x s2. Thus, f is decreasing on the interval (s2, ). This means that f n 1
fn
and therefore bn 1 bn when n 2. (The inequality b2 b1 can be veriﬁed directly but
all that really matters is that the sequence bn is eventually decreasing.)
Condition (ii) is readily veriﬁed: lim bn n lim nl n3 nl 1
n 2 lim 1 nl 1
n3 1 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 sn is the remainder Rn s 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 bn 1 , which is the absolute value of the ﬁrst neglected term.
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
b6, and so on. Notice
s s4 b 5, s s5
also that s lies between any two consecutive
partial sums. 1 n1 bn is the sum of an alternating series that satisﬁes
(i) 0 bn then and bn 1 Rn s (ii) lim bn
nl sn bn 0 1 Proof We know from the proof of the Alternating Series Test that s lies between any two
consecutive partial sums sn and sn 1 . It follows that s sn sn EXAMPLE 4 Find the sum of the series (By deﬁnition, 0! n0 1.) sn 1 1
n! bn 1 n correct to three decimal places. SOLUTION We ﬁrst observe that the series is convergent by the Alternating Series Test because
(i) (ii) 1 1
n
0 1!
1
n! n! n 1 1
n! 1
1
l 0 so
l 0 as n l
n
n! 5E12(pp 766775) 1/18/06 10:55 AM Page 775 S ECTION 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
0!
1 1
1! 1
2!
1
2 1 1
3! 1
6 1
24 s6 and 1 1
2 1 1
120 1
5040 b7 Notice that 1
4! 1
5!
1
720 1
5000 1
6 1
24 1
6! 1
7! 1
5040 0.0002
1
120 1
720 0.368056 By the Alternating Series Estimation Theorem we know that
s s6 b7 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
n
ex
n 0 x n! for all x, so what we have
obtained in Example 4 is actually an approximation to the number e 1. s 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? (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. 1
3 4. n1 cos n
n3 4 16.
n1 1 n sin 17.
n1 2–20 3. 15. 4
7 Test the series for convergence or divergence.
2
4 3
5 4
8 4
9 1
ln 2 5.
n1 7. 1n
sn
n 1 n1 11. 1
n2 1
ln 4 ■ 1
ln 5 1
ln 6
1 6.
n1 1
1 8. 3n
1 1 n2 n 4n 2 n
n n 1 n
ln n 4 12. n1 1 n1 14. 1
n1 21. 2n 1
n1 3 ■ nn
n! ■ 20.
n1
■ ■ ■ ■ ■ n
5 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. n1 n1 10. n1 n n n1 ; 21–22 1 3n
2n 1
n1 4
11 1 n1 13. 4
10 n1 1
4n 2
1 19. 5
7 1
ln 3 n1 9. 4
6 1 n cos 18. n sin n 2
n! n1 1 sn
2 sn n1
■ ■ 1
n3 n1 22. 2 n1 ■ ■ ■ ■ 23–26 ■ ■ 1n
n3
■ 1 ■ ■ ■  How many terms of the series do we need to add in order
to ﬁnd the sum to the indicated accuracy? e1 n
n 23. ln n
n 24. 1 n1 1n
n2 1 n1 1n
n4 ( error 0.01) ( error 0.001) 5E12(pp 776785) 776 ❙❙❙❙ 1/18/06 10:58 AM Page 776 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2
n! 25.
n1 n ( 0.01) error 34. 1
n2 ■ 1 nn
4n 26.
n1
■ ■ 27–30 ( ■  ■ error
■ 0.002)
■ n1 ■ ■ ■ 1 ■ ■ ■ n1 1
n
10 n 29.
n1
■ ■ ■ 2 30.
n1
■ ■ ■ ■ ■ ■ ■ ■ ■ hn 1 n 1 n an overestimate or an underestimate of the
total sum? Explain. 32.
n1 1  12.6 ■ 1 33.
n1 n ■ ■ ■ ■ ■ n1 1n
n 1 ln 2 as n l ln n l and therefore
h2 n For what values of p is each series convergent?
1n
np ■ Let hn and sn be the partial sums of the harmonic and alternating harmonic series.
(a) Show that s2 n h2 n hn.
(b) From Exercise 38 in Section 12.3 we have 1n
n
3 n! n1  ■ n1 31. Is the 50th partial sum s50 of the alternating series 32–34 ■ 36. Use the following steps to show that 1 nn
8n 28. n1 ■ p 1 bn , where bn 1 n if n is odd
and bn 1 n 2 if n is even, is divergent. Why does the Alternating Series Test not apply? Approximate the sum of the series correct to four decimal
1n
n5 ■ ln n
n 35. Show that the series places.
27. n1 as n l ln 2 n l Use these facts together with part (a) to show that
s2 n l ln 2 as n l . n p Absolute Convergence and the Ratio and Root Tests
Given any series a n , we can consider the corresponding series
an a1 a2 a3 n1 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. Notice that if a n is a series with positive terms, then a n
vergence is the same as convergence in this case. a n and so absolute con EXAMPLE 1 The series 1n
n2 n1 1 1 1
22 1
32 1
42 1
n2 1 1
22 1
32 is absolutely convergent because n1 1n
n2 is a convergent pseries ( p 1
n1 2). 1
42 5E12(pp 776785) 1/18/06 10:58 AM Page 777 S ECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 777 EXAMPLE 2 We know that the alternating harmonic series 1n
n n1 1 1 1
2 1
3 1
4 is convergent (see Example 1 in Section 12.5), but it is not absolutely convergent
because the corresponding series of absolute values is
1n
n n1 1 1
n n1 1
2 1 which is the harmonic series ( pseries with p 1
3 1
4 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.
3 Theorem If a series a n is absolutely convergent, then it is convergent. Proof Observe that the inequality 0 an an 2 an is true because a n is either a n or a 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
an )
is convergent. Then (a n an an ) an 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 n1 cos n
n2 cos 1
12 cos 2
22 cos 3
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
cos n
cos n
n2
n2
n1
n1 Since cos n 1 for all n, we have
cos n
n2 1
n2 5E12(pp 776785) 778 ❙❙❙❙ 1/18/06 10:58 AM Page 778 CHAPTER 12 INFINITE SEQUENCES AND SERIES We know that 1 n 2 is convergent ( pseries with p 2) and therefore
cos n n 2 is
2
convergent by the Comparison Test. Thus, the given series cos n n is absolutely
convergent and therefore convergent by Theorem 3.
The following test is very useful in determining whether a given series is absolutely
convergent.
The Ratio Test an 1
L 1, then the series
nl
an
n
(and therefore convergent). (i) If lim an 1
an
is divergent. (ii) If lim an 1
an 1 or lim L nl nl a n is absolutely convergent
1 , then the series an
n1 an 1
1, the Ratio Test is inconclusive; that is, no conclusion can
nl
an
be drawn about the convergence or divergence of a n. (iii) If lim Proof L (i) The idea is to compare the given series with a convergent geometric series. Since
1, we can choose a number r such that L r 1. Since
lim nl the ratio a n
such that 1 an 1
an L and L r a n will eventually be less than r ; that is, there exists an integer N
an 1
an r whenever n N or, equivalently,
an 4 an r 1 Putting n successively equal to N , N whenever n 1, N N 2, . . . in (4), we obtain aN 1 aN r aN 2 aN 1 r aN r 2 aN 3 aN 2 r aN r 3 and, in general,
aN 5 k aN r k for all k 1 Now the series
aN r k
k1 aN r aN r 2 aN r 3 5E12(pp 776785) 1/18/06 10:58 AM Page 779 S ECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS is convergent because it is a geometric series with 0 r
together with the Comparison Test, shows that the series
an aN nN1 aN k aN 1 ❙❙❙❙ 779 1. So the inequality (5), aN 2 3 k1 is also convergent. It follows that the series n 1 a n is convergent. (Recall that a ﬁnite
number of terms doesn’t affect convergence.) Therefore, a n is absolutely convergent.
(ii) If a n 1 a n l L 1 or a n 1 a n l , then the ratio a n 1 a n will eventually be
greater than 1; that is, there exists an integer N such that
an 1
an
This means that a n 1 whenever n a n whenever n 1 N and so lim a n 0 nl Therefore,
NOTE N a n diverges by the Test for Divergence. Part (iii) of the Ratio Test says that if lim n l a n 1 a n
information. For instance, for the convergent series 1 n 2 we have
■ 1, the test gives no 1
an 1
an n 2 1 n2 1
n2 n whereas for the divergent series 1
1 2 1 l1 2 1
n as n l 1 n we have 1
an 1
an n 1 n 1
n n 1
1 l1 1
n 1 as n l 1, the series a n might converge or it might diverge. In
Therefore, if lim n l a n 1 a n
this case the Ratio Test fails and we must use some other test.
EXAMPLE 4 Test the series 1 n n1  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
1 n n 3 3 n: SOLUTION We use the Ratio Test with a n 
1
3 an 1
an 1 n1 n n 1 3 3n 1
1 nn 3
3n
1
n 3 1
3  n 1
3 1 3 3n
n3 n1 1
n 3 l 1
3 1 Thus, by the Ratio Test, the given series is absolutely convergent and therefore
convergent. 5E12(pp 776785) 780 ❙❙❙❙ 1/18/06 10:58 AM Page 780 CHAPTER 12 INFINITE SEQUENCES AND SERIES E XAMPLE 5 Test the convergence of the series
n1
n SOLUTION Since the terms a n an 1
an n n! are positive, we don’t need the absolute value signs.
1n 1
1! n
n
n 1 n!
nn n 1 (See Equation 7.4.9 or 7.4*.9.) Since e
Ratio Test. n 1n 1
n 1 n! 1
n n n NOTE nn
.
n! n le n!
nn as n l 1, the given series is divergent by the Although the Ratio Test works in Example 5, an easier method is to use the Test
for Divergence. Since
■ nn
n! an nnn
123 n
n n it follows that a n does not approach 0 as n l . 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
n
(i) If lim s a n nl L 1, then the series a n is absolutely convergent
n1 (and therefore convergent).
n
(ii) If lim s a n nl n
(iii) If lim s a n nl L n
1 or lim s a n a n is divergent. , then the series nl n1 1, the Root Test is inconclusive. n
1, then part (iii) of the Root Test says that the test gives no inforIf lim n l s a n
mation. 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.) EXAMPLE 6 Test the convergence of the series
n1 2n
3n 3
2 n . SOLUTION an n
s an 2n
3n
2n
3n 3
2
3
2 n 2
3 Thus, the given series converges by the Root Test. 3
n
2
l
3
2
n 1 5E12(pp 776785) 1/18/06 10:59 AM Page 781 S ECTION 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:
a1 a2 a5 a3 a4 a 15 a6 a7 a 20 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
2 1 6 1
3 1
4 1
5 1
6 1
7 1
8 ln 2
1
2 (See Exercise 36 in Section 12.5.) If we multiply this series by , we get
1
2 1
4 1
6 1
8 1
2 ln 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 7 1
2 0 1
4 1
6 0 0 1
8 1
2 ln 2 Now we add the series in Equations 6 and 7 using Theorem 12.2.8:
1 8 1
3 1
2 1
5 1
7 1
4 3
2 ln 2 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
nl n0 an 1
an an 1
(c) lim
nl
an
2–28 8 (b) lim
nl an 1
an 2.
n1 n2
2n 10
n! n 4. 1 0.8 1 5.
n1 7. 1
sn n1 11.
n1 n1 n n n1 9. 6. 4 1 n1 n1 n1 Determine whether the series is absolutely convergent,
conditionally convergent, or divergent.
 3. 5 1
2n !
1 ne 1 n
n3 n 8. n 1
n4
1 e n n!
n1 12.
n1 n n1 n1 10. 2n
n4 sin 4 n
4n n2 1 5E12(pp 776785) 782 ❙❙❙❙ n1 15. 3
4 14. n2 n1 n1 cos n 3
n! n1 23.
n1 31 2
5 n1 n an 1 1 ■ ■ the series ■ 2
■ 37. Prove that if
■ ■ ■ ■ a n is absolutely convergent, then ■ an 29. The terms of a series are deﬁned recursively by the equations a1 an 2 Determine whether
30. A series 5n
4n 1 1
an
3 a n converges or diverges. a n is deﬁned by the equations
a1 an 1 Determine whether 2
1 cos n
an
sn a n converges or diverges. n1 (that is, it fails to give a deﬁnite answer)?
1
n
(a)
(b)
3
n
n1n
n12 n1 3n
sn 1 (d)
n1 1 sn
n2 32. For which positive integers k is the following series convergent? n1 33. (a) Show that n n! 2
kn ! n 0 x n! converges for all x.
(b) Deduce that lim n l x n n! 0 for all x. an
n1 38. Prove the Root Test. [Hint for part (i): Take any number r
1 and use the fact that there is an integer N
r whenever n N .] such that L r
n
such that s a n
39. Given any series a n we deﬁne a series a n whose terms are
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
an 31. For which of the following series is the Ratio Test inconclusive (c) n
2n Use Exercise 34 to estimate the error.
3n ■ n1 36. Use the sum of the ﬁrst 10 terms to approximate the sum of 2 n n! ■ 1 1 n2 n. Use Exercise 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. 2 6 10 14
5 8 11 14 5 8 11 n1 1 rn 35. (a) Find the partial sum s5 of the series n1 n 1, show, by 1 an 1
1L Rn
n n!
1 3 (b) If rn is an increasing sequence, show that 2n n1 an 2 an Rn 135
1357
5!
7!
135
2n 1
n1
1
2n 1 !
2 6 10
5 8 11 an (a) If rn is a decreasing sequence and rn
summing a geometric series, that n 1n
arctan n 24. 246 28. Rn 1n
n ln n n 1
1 26
58 27. ■ n2 n2 13
3! 25. 1 1
ln n 22. n2
2n 2 a n be a series with positive terms and let rn a n 1 a n.
Suppose that lim n l rn L 1, so a n converges by the
Ratio Test. As usual, we let Rn be the remainder after n terms,
that is, n!
nn 20. 3n 34. Let 3 cos n
n2 3 2 18. nn 21. n 22n
n! n1 16. 1 1n
ln n 19. n1 1
n1 10 n
1 42n n 17. Page 782 n n1 n1 26. 10:59 AM CHAPTER 12 INFINITE SEQUENCES AND SERIES n 13. 1/18/06 an an
2 an an an
2 Notice that if a n 0, then a n a n and a n
0, whereas if
a n 0, then a n a n and a n 0.
(a) If a n is absolutely convergent, show that both of the series
a n and a n are convergent.
(b) If a n is conditionally convergent, show that both of the
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 a n so that their sum is greater
than r . Then add just enough negative terms a n so that the
cumulative sum is less than r . Continue in this manner and use
Theorem 12.2.6.] 5E12(pp 776785) 1/18/06 10:59 AM Page 783 S ECTION 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 n p, it is a pseries, which we know to be convergent if p 1 and divergent if p 1.
2. If the series has the form ar n 1 or ar n, it is a geometric series, which converges if r
1 and diverges if r
1. Some preliminary algebraic manipulation may be required to bring the series into this form.
1. If the series is of the form 3. If the series has a form that is similar to a pseries or a geometric series, then 4.
5.
6. 7.
8. 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 pseries. 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 a n 0, then the Test for Divergence
should be used.
If the series is of the form
1 n 1bn or
1 nbn , 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 n 1 a n l 1 as n l for all pseries 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 n , where x1 f x d x is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisﬁed). In the following examples we don’t work out all the details but simply indicate which
tests should be used. EXAMPLE 1
n1 Since a n l 1
2 EXAMPLE 2
n1 n
2n 1
1 0 as n l , we should use the Test for Divergence.
sn 3 1
3n 3 4n 2 2 Since a n is an algebraic function of n, we compare the given series with a pseries. 5E12(pp 776785) ❙❙❙❙ 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 EXAMPLE 3 ne sn 3
3n 3 n3 2
3n 3 bn , where 1
3n 3 2 n2 n1 Since the integral x1 xe
also works.
EXAMPLE 4 1
n1 x2 dx is easily evaluated, we use the Integral Test. The Ratio Test n3 n n4 1 Since the series is alternating, we use the Alternating Series Test.
EXAMPLE 5
k1 2k
k! Since the series involves k!, we use the Ratio Test.
1 EXAMPLE 6
n1 2 3n Since the series is closely related to the geometric series
Test.  12.7
1–38  1.
n1 Exercises Test the series for convergence or divergence.
n2
n2 1
n 2.
n1 1 3.
n1 n 2 n1 7.
n2 n
n2 4. n
3n
2 3n 5. n1 1 1 6.
n1 8.
k1 19. 1
n n1 1
n s ln n 1 k n
n2 1
n tan 1 n k1 25.
n1 2 k k!
k 2! 27. n3 29. n1 n1 11.
n2 1n1
n ln n
n 13.
n1 3n
n! 12. n0 258
1 n21 n 17.
n1 n2 25 14. 3n 2 n1 n2 1n 1
sn 1 18. 1
1 n1 n (s2 37. n n1 e1 n
n2
1 32. 4k n1 34.
n1
n2 36. 1 n2
n 1) j n
1
ln n
n n1 5 2n n
n 2n (s2 38. sj j j1 n
n1 1
5n 30. sin 1 n
sn 35. n2 28. tan 1n
n sn
3k 33. n2
n3 16. k ln k
k 13 n1 sin n 3 n1 k1 n1 cos n 2
n 2 4n 26. 5k 31. sn 2 1
n
2n 2 5 n1 n!
2
en n n 2 n! 15. 1
n1 24. n1 n1 n 5
5k k1 22. n1 23. k 20. 2 2n
nn 21. 3n
1 8n n 2e 10. ln n
sn n n1 k1 k 2e 9. 1 3 n, we use the Comparison 1
n cos2 n
ln n 1) 5E12(pp 776785) 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
cn x n 1 c0 c2 x 2 c1 x c3 x 3 n0 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
a n cos n x bn sin n x fx c0 c2 x 2 c1 x cn x n 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 n0 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. xn 1 x x2 xn n0 which converges when 1 x 1 and diverges when x
More generally, a series of the form
cn x 2 a n c0 c1 x a 1 (see Equation 12.2.5). c2 x a 2 n0 is called a power series in x a 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 a 0 1 even when x a. Notice also that when
x a all of the terms are 0 for n 1 and so the power series (2) always converges when
x a.
n! x n convergent? EXAMPLE 1 For what values of x is the series
n0 SOLUTION We use the Ratio Test. If we let a n , as usual, denote the n th term of the series, then a n n! x n. If x 0, we have
an 1
an lim nl n lim nl 1 !xn
n! x n By the Ratio Test, the series diverges when x
when x 0. 1 lim n nl 0. Thus, the given series converges only
x EXAMPLE 2 For what values of x does the series
n1 SOLUTION Let a n x 3 n 1x 3
n n converge? n. Then an 1
an x 3
n 1 1
1 n1 1
n x n
x 3 n 3lx 3 as n l 5E12(pp 786795) 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 3
1 and divergent when x 3
1. Now
x 3 1 1 &? x 3 1 2 &? x 4 so the series converges when 2 x 4 and diverges when x 2 or x 4.
The Ratio Test gives no information when x 3
1 so we must consider x 2
and x 4 separately. If we put x 4 in the series, it becomes 1 n, the harmonic
series, which is divergent. If x 2, the series is
1 n n, which converges by the
Alternating Series Test. Thus, the given power series converges for 2 x 4.
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 1 nx 2n
2 n! 2 J0 x 2n n0 1 n x 2 n 2 2 n n! 2 . Then SOLUTION Let a n an 1
an 1
2
2 n 1 2n 1 2n 2 n 1! x 2n 2
n 1 2 n! x2
4n 2 2 n n! 2
1 nx 2 n x 2n 1 1 2 2 2 2 n n!
x 2n 2 l0 1 2 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
,
.
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 J0 x lim sn x where nl 1 ix 2i
2 i! 2 sn x 2i i0 The ﬁrst few partial sums are
s0 x s3 x 1 x2
4 1 s1 x
x4
64 x6
2304 1 x2
4
s4 x s2 x 1 x2
4 1
x4
64 x2
4 x4
64
x6
2304 x8
147,456 5E12(pp 786795) 1/18/06 10:36 AM Page 787 S ECTION 12.8 POWER SERIES y s™ s¸ 1 s¢
0 x 1 s¡ s£ 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
,
in Example 3,
and a collapsed interval 0, 0
0 in Example 1]. The following theorem, proved in
Appendix F, says that this is true in general. 3 Partial sums of the Bessel function J¸
y
1 y=J¸(x) 10
0 787 J¸ FIGURE 1 _10 ❙❙❙❙ x FIGURE 2 Theorem For a given power series cn x a n there are only three n0 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
and diverges if x a
R. a R 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
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
, . In case (iii) note that the inequality x a
R can be rewritten as a R x a R. When x is an endpoint of the interval, that is, x a 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:
a R, a R a R, a R a R, a R a R, a R The situation is illustrated in Figure 3.
convergence for  xa <R
aR FIGURE 3 a+R a divergence for  xa >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 xn R 1 1, 1 n! x n Geometric series R 0 0 R 1 2, 4 n0 Example 1
n0 x Example 2 3 n n1 n n0 1 nx 2n
2 2 n n! 2 Example 3 R , 5E12(pp 786795) 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 3 nx n
sn 1 n0 3 n x n sn SOLUTION Let a n 1. Then 3 n 1x n
sn 2 an 1
an 1
1 3 1 sn 1
3 nx n 1n
2n 3x x l3x n
n 1
2 as n l By the Ratio Test, the given series converges if 3 x
1 and diverges if 3 x
1.
1
1
Thus, it converges if x
3 and diverges if x
3 . This means that the radius of
convergence is R 1 .
3
We know the series converges in the interval ( 1 , 1 ), but we must now test for
33
1
convergence at the endpoints of this interval. If x
3 , the series becomes
n n0 3 n( 1 )
3
sn 1 1
n0 sn 1
s1 1 1
s2 1
s3 1
s4 which diverges. (Use the Integral Test or simply observe that it is a pseries with
p 1 1.) If x 1 , the series is
2
3 n0 3
sn n1n
3 () 1 n 1
n0 sn 1 which converges by the Alternating Series Test. Therefore, the given power series converges when 1 x 1 , so the interval of convergence is ( 1 , 1 ].
3
3
33
EXAMPLE 5 Find the radius of convergence and interval of convergence of the series n0 SOLUTION If a n nx
an 1
an 2 n nx 2
3n 1 n 3 n 1, then
n 1 1x 2
3n 2
1
n x n1 2
3 3n 1
nx 2
l x n 2
3 as n l Using the Ratio Test, we see that the series converges if x 2 3 1 and it diverges if
x 2 3 1. So it converges if x 2
3 and diverges if x 2
3. Thus, the
radius of convergence is R 3. 5E12(pp 786795) 1/18/06 10:38 AM Page 789 ❙❙❙❙ S ECTION 12.8 POWER SERIES 3 can be written as 5
The inequality x 2
5, the series is
the endpoints 5 and 1. When x
n
n0 3
3n 1, so we test the series at n
1
3 1 1 nn
n0 1 nn doesn’t converge to 0]. When x which diverges by the Test for Divergence [
the series is
n3n
3n 1 n0 x 789 1
3 1, n
n0 which also diverges by the Test for Divergence. Thus, the series converges only when
5 x 1, so the interval of convergence is 5, 1 .  12.8 Exercises 1. What is a power series?
2. (a) What is the radius of convergence of a power series? 27.
n2 28. Find the radius of convergence and interval of convergence
of the series.
3.
n1 n1 7.
n0 4.
n0 1 n 1xn
n3 5. 1 n 246
135
■ 2n
xn
2n 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ n n 0 cn 4 is convergent, does it follow that the following
series are convergent? cn (a) 2 n cn (b) n0 4 n n0 n
4 and diverges
n 0 cn x converges when x
when x 6. What can be said about the convergence or divergence of the following series? xn
n3n (a)
(c) n1 xn
5nn5 10.
12. xn
4 ln n 14. n n2 sn x 1 1 n n3 x 16. 17. 1 x n n1 x 2
n 2
n2 n 18.
n1 n 20. n n1 n
x
bn a, n! 2 x 1 n b 0 1 ncn 9 n (d)
n0 n0 n! k n
x
kn ! ; 32. Graph the ﬁrst several partial sums sn x of the series
2
sn x 3x 2
n3n 3 n n x n, 33. The function J1 deﬁned by 4n
1 n1 n0 together with the sum function f x
1 1 x , on a
common screen. On what interval do these partial sums appear
to be converging to f x ? n n2xn
246 24. n n 5 nx
n3 22. 3 the series n1 n cn 31. If k is a positive integer, ﬁnd the radius of convergence of x 2n
2n ! n0
n n0 n0 n cn 8 n (b) cn
n0 n0 n0 n1 n n 29. If 1x
1 n1 2 nx n
sn 13. 23. n2 2x 3
n ln n 30. Suppose that 4 n1 n1 n 1 n1 11. 21. 26. n nx n 8. n1 n1 n x
ln n ■ n1 1 nn 4 n x n 19. ■ sn x n 6. xn
n! 9. 15. n1 nn x
sn n 2 n  n 1
n n1 How do you ﬁnd it?
(b) What is the interval of convergence of a power series?
How do you ﬁnd it?
3–28 4x 25. J1 x
n0 2n ; 1 nx 2n 1
n! n 1 ! 2 2 n 1 is called the Bessel function of order 1.
(a) Find its domain.
(b) Graph the ﬁrst several partial sums on a common screen. 5E12(pp 786795) ❙❙❙❙ 790 CAS 1/18/06 10:39 AM Page 790 CHAPTER 12 INFINITE SEQUENCES AND SERIES that is, its coefﬁcients are c2 n 1 and c2 n 1 2 for all n
Find the interval of convergence of the series and ﬁnd an
explicit formula for f x . (c) If your CAS has builtin Bessel functions, graph J1 on the
same screen as the partial sums in part (b) and observe how
the partial sums approximate J1. n
cn for all n 0, ﬁnd the
n 0 cn x , where cn 4
interval of convergence of the series and a formula for f x . 36. If f x 34. The function A deﬁned by Ax ;
CAS x3
23 1 x6
2356 x9
235689 n
37. Show that if lim n l s cn c, where c 0, then the radius
of convergence of the power series cn x n is R 1 c. 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 sn x on a common
screen.
(c) If your CAS has builtin Airy functions, graph A on the
same screen as the partial sums in part (b) and observe how
the partial sums approximate A. cn x a n satisﬁes c n 0 for
all n. Show that if lim n l cn cn 1 exists, then it is equal to
the radius of convergence of the power series. 38. Suppose that the power 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 dn x n ? 39. Suppose the series 40. Suppose that the radius of convergence of the power series 35. A function f is deﬁned by fx  12.9 1 2x 0. x 2 2x 3 x cn x n is R. What is the radius of convergence of the power
series cn x 2 n ? 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 sn x
nl
1x
where
sn x 1 x x2 1 1
1 x 1 x x2 x3 xn x 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 x
1 1 x as a sum
of a power series. xn s¡¡ y is the n th partial sum. Notice that as n
increases, sn x becomes a better approximation
to f x for 1 x 1. sˆ
s∞
f s™
FIGURE 1 1
ƒ=
and some partial sums
1x 1 n0 _1 0 1 x 5E12(pp 786795) 1/18/06 10:39 AM Page 791 S ECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 791 x 2 as the sum of a power series and ﬁnd the interval of EXAMPLE 1 Express 1 1 convergence.
x 2 in Equation 1, we have SOLUTION Replacing x by 1 1
x x2 x2 1 n n0 1 nx 2n 1 2 1 x2 x4 x6 x8 n0 Because this is a geometric series, it converges when
1, that is, x 2 1, or
x2
1. Therefore, the interval of convergence is 1, 1 . (Of course, we could have
x
determined the radius of convergence by applying the Ratio Test, but that much work is
unnecessary here.)
EXAMPLE 2 Find a power series representation for 1 x 2. 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 1 2 x This series converges when
is 2, 2 . x
2 21
1
2 1 x
2 n0 x
2 21
n
n0 2 1, that is, x x2 n 1
n1 xn 2. So the interval of convergence EXAMPLE 3 Find a power series representation of x 3 x 2. 3 SOLUTION Since this function is just x 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.] x3
x 2 x3
1
2 x3 1
x 2
1
4 2n n0
1
8 x4 n 1 x3
x5 1
16 1 n 1 xn
n0 2n 1 xn 3 x6 Another way of writing this series is as follows:
x3
x 2 n3 As in Example 2, the interval of convergence is 1
2n n1
2 xn 2, 2 . Differentiation and Integration of Power Series
The sum of a power series is a function f x
a n whose domain is the intern 0 cn x
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 termbyterm differentiation and integration. 5E12(pp 786795) 792 ❙❙❙❙ 1/18/06 10:40 AM Page 792 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2 Theorem If the power series a n has radius of convergence R cn x 0, then the function f deﬁned by
fx c0 c1 x a c2 x a 2 cn x a n n0 is differentiable (and therefore continuous) on the interval a
(i) f x c1 2c2 x a 3c3 x 2 a R, a ncn x a R and
n1 n1  In part (ii), x c0 dx c0 x C1 is written
C, where C C1 ac0, so
as c0 x a
all the terms of the series have the same form. (ii) yf x dx C c0 x C a cn x a
n n0 x c1 a 2 c2 2 x a 3 3 n1 1 The radii of convergence of the power series in Equations (i) and (ii) are both R.
NOTE 1 (iii)
(iv) ■ d
dx Equations (i) and (ii) in Theorem 2 can be rewritten in the form
cn x a n n0 y cn x n0 a n n0 d
cn x
dx yc dx n a n a n dx x n0 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 J0 x
n0 1 nx 2n
2 2 n n! 2 is deﬁned for all x. Thus, by Theorem 2, J0 is differentiable for all x and its derivative is
found by termbyterm differentiation as follows:
J0 x
n0 EXAMPLE 5 Express 1 1 the radius of convergence? d
1 nx 2 n
2n
dx 2 n! 2 n1 1 n 2nx 2 n
2 2 n n! 2 1 x 2 as a power series by differentiating Equation 1. What is 5E12(pp 786795) 1/18/06 10:41 AM Page 793 S ECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 793 SOLUTION Differentiating each side of the equation 1
1 1 x 1 1 2 x x3 xn
n0 1 we get x2 x 3x 2 2x nx n 1 n1 If we wish, we can replace n by n 1 and write the answer as 1
1 1 xn n 2 x n0 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 ln 1
convergence. x and its radius of SOLUTION We notice that, except for a factor of 11 1, the derivative of this function is
x . So we integrate both sides of Equation 1:
ln 1 y1 x 1
x
x2
2 x n1 x xn
n x2 x dx C
n0 C x xn 1
n1 1 x3
3 n1 xn
n 1
2 1
2 ln 2 1
8 1
24 1
64 n1 tan 1x y
C C. 1.
ln 2, we 1
n2n tan 1x. 1 1 x 2 and ﬁnd the required series by integrating
found in Example 1. SOLUTION We observe that f x x 0 1 in the result of Example 6. Since ln 1
2 EXAMPLE 7 Find a power series representation for f x the power series for 1 1 ln 1 x The radius of convergence is the same as for the original series: R
Notice what happens if we put x
see that C 0 in this equation and obtain x2
2 x 1 x3
3 To determine the value of C we put x
Thus, C 0 and
ln 1 y dx 2 1
x2 1
x dx y x3
3 x5
5 1 x2
x7
7 x4 x6 dx 5E12(pp 786795) 794 ❙❙❙❙ 1/18/06 10:42 AM Page 794 CHAPTER 12 INFINITE SEQUENCES AND SERIES  The power series for tan 1x 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 1 x 1, but it turns
out (although it isn’t easy to prove) that it is also
valid when x
1. Notice that when x 1
the series becomes
4 1 1
3 1
5 1
7 This beautiful result is known as the Leibniz
formula for . To ﬁnd C we put x tan 1 0 0 and obtain C tan 1x x3
3 x x5
5 0. Therefore x7
7 1 n n0 x 2n 1
2n 1 x 2 is 1, the radius of conver Since the radius of convergence of the series for 1 1
gence of this series for tan 1x is also 1.
EXAMPLE 8 (a) Evaluate x 1 1 x 7 d x as a power series.
(b) Use part (a) to approximate x00.5 1 1 x 7 d x correct to within 10 7.
SOLUTION (a) The ﬁrst step is to express the integrand, 1 1 x 7 , as the sum of a power series.
As in Example 1, we start with Equation 1 and replace x by x 7:
1 1
x x7 x7 1 n n0 1 n x 7n 1 7 1 x7 x 14 n0  This example demonstrates one way in
which power series representations are useful.
Integrating 1 1 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: y1 1
x7 1 n x 7n d x y dx C 1 n0 C n0 x8
8 x x 15
15 n x 7n 1
7n 1 x 22
22 This series converges for
x7
1, that is, for x
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 1 0.5 0 1 x7 dx x
1
2 x8
8
1
8 28 x 15
15 12 x 22
22 0 1
15 2 1
15 22 2 22 7n 1n
1 2 7n 1 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
29 2 29 6.4 10 11 So we have y 1 0.5 0 1 x7 dx 1
2 1
8 28 1
15 2 15 1
22 2 22 0.49951374 5E12(pp 786795) 1/18/06 10:43 AM Page 795 SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES  12.9 n0 cn x n 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
n1
? Why?
n 1 ncn x
2. Suppose you know that the series 15. f x n bn x converges for
2. What can you say about the following series? Why? n0 n n0 x 1 5. f x 7. f x 9. f x
■ ■ 11–12 1
1
1
x
x
9
■ ■ ■ ■ ■ ■ 19. f x
21. f x ln x ■ 1
x a3
■ x3
■ ■ Express the function as the sum of a power series by ﬁrst
using partial fractions. Find the interval of convergence.
 11. f x
12. f x
■ ■ x2 2 7x
■ ■ ■ ■ ■ ■ 1
1 x
x ■ 1 20. f x x x2 22. f x ■ ■ 1 tan ■ ■ 25 ■ ■ 2x
■ ■ ■ 23. y1 25. y t
t8 x tan
x3 ■ 1 ■ ln 1 24. x dx
■ ■ ■ y 26. dt y tan t dt t ■ ■ 1 x 2 dx
■ ■ ■ ■  Use a power series to approximate the deﬁnite integral to
six decimal places. 2x
■ ■ ■ 27–30 1 3x 2 ■ ■ 3
x arctan x 3 23–26  Evaluate the indeﬁnite integral as a power series. What is
the radius of convergence? 2 ■ 18. f x
■ ln 3 4x 2 ■ 9x 2 1 10. f x x2 ■ 1 8. f x 5 ■ 2 x4 1 6. f x x3 1 x 2  Find a power series representation for f , and graph f and
several partial sums sn x on the same screen. What happens as n
increases? 3 4. f x x x2
1 2x ; 19–22 3–10  Find a power series representation for the function and
determine the interval of convergence. 1 16. f x x3 n1
■ x ln 5 17. f x bn 3. f x 795 Exercises 1. If the radius of convergence of the power series x ❙❙❙❙ 1
■ ■ ■ ■ ■ ■ ■ ■ 13. (a) Use differentiation to ﬁnd a power series representation for fx 1
1 x 29. y 1 0.2 13 0
■ x 2 tan
■ 28. 1 x 4 dx
■ ■ ■ y 30. dx x5 1 0 y ■ x x6 1
■ ■ ■ 32. Show that the function fx 3 n0 3 14. (a) Find a power series representation for f x What is the radius of convergence?
(b) Use part (a) to ﬁnd a power series for f x
(c) Use part (a) to ﬁnd a power series for f x 1 nx 2n
2n ! is a solution of the differential equation
fx x dx 0.5 0 decimal places. x2
1 x 4 dx ln 1 ■ 31. Use the result of Example 6 to compute ln 1.1 correct to ﬁve 1
1 0.4 0 2 (c) Use part (b) to ﬁnd a power series for
fx y ■ What is the radius of convergence?
(b) Use part (a) to ﬁnd a power series for
fx 27. fx 0 33. (a) Show that J0 (the Bessel function of order 0 given in ln 1 x. x ln 1
ln x 2 x.
1. Example 4) satisﬁes the differential equation
x 2J 0 x x J0 x x 2J0 x 0 (b) Evaluate x01 J0 x d x correct to three decimal places. ■ 5E12(pp 796805) 796 ❙❙❙❙ 1/18/06 10:29 AM Page 796 CHAPTER 12 INFINITE SEQUENCES AND SERIES 38. (a) Starting with the geometric series 34. The Bessel function of order 1 is deﬁned by n0 x n, ﬁnd the sum of the series
J1 x
n0 1 nx 2n 1
n! n 1 !2 2 n nx n 1 x J1 x (b) Show that J0 x x2 1 J1 x 0 (i) 35. (a) Show that the function 1 x n, nn x 1 n2 fx
n0 xn
n! n2 (ii) n
2 n2 (iii) n n1 39. Use the power series for tan is a solution of the differential equation n2
2n 1 x to prove the following expresas the sum of an inﬁnite series: sion for
(b) Show that f x 1 (c) Find the sum of each of the following series. J1 x . fx x (b) Find the sum of each of the following series.
n
(i)
(ii)
n x n, x
1
2n
n1
n1 (a) Show that J1 satisﬁes the differential equation
x 2J1 x 1 n1 fx
2s3 x e. n0 1n
2n 1 3 n 40. (a) By completing the square, show that sin n x n 2. Show that the series fn x converges
for all values of x but the series of derivatives f n x diverges
when x 2 n , n an integer. For what values of x does the
series f n x converge? 36. Let fn x y dx
x 12 x2 0 1 3s3 3 (b) By factoring x
1 as a sum of cubes, rewrite the integral
in part (a). Then express 1 x 3 1 as the sum of a power
series and use it to prove the following formula for : 37. Let fx
n1 xn
n2 3 s3
4 Find the intervals of convergence for f , f , and f .  12.10 1
8n n0 n 2
3n 1
1 3n 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 fx c0 c1 x a c2 x a 2 c3 x a 3 c4 x a 4 x a 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
fa c0 By Theorem 12.9.2, we can differentiate the series in Equation 1 term by term:
2 fx c1 2c2 x and substitution of x a 3c3 x a 2 4c4 x a in Equation 2 gives
fa c1 a 3 x a R 5E12(pp 796805) 1/18/06 10:30 AM Page 797 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 797 Now we differentiate both sides of Equation 2 and obtain
3 fx 2c2 2 3c3 x Again we put x a 3 4c4 x a 2 x a R a in Equation 3. The result is
fa 2c2 Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives
4 f x 2 3c3 2 3 4c4 x and substitution of x a 3 4 5c5 x a 2 x a R a in Equation 4 gives
fa 2 3c3 3! c3 By now you can see the pattern. If we continue to differentiate and substitute x
obtain
n f a 234 n cn a, we n! cn Solving this equation for the nth coefﬁcient cn , we get
f cn n a n! This formula remains valid even for n 0 if we adopt the conventions that 0!
f0
f . Thus, we have proved the following theorem.
5 1 and Theorem If f has a power series representation (expansion) at a, that is, if fx cn x a n x a R n0 then its coefﬁcients are given by the formula
cn f n a 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 fx
n0 fa n n! a x fa
1! a
x n a fa
2! x a 2 fa
3! x a 3 5E12(pp 796805) 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 fx 7 n 0 xn n! n0 f0
x
1! f0 f0 2
x
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 x e x and its radius of convergence.
e x, then f n x
e x, so f n 0
e0
Taylor series for f at 0 (that is, the Maclaurin series) is
SOLUTION If f x f n 0 n! n0 xn
n0 xn
n! xn
n 1 1! x3
3! x n n!. Then To ﬁnd the radius of convergence we let a n
an 1
an x2
2! x
1! 1 1 for all n. Therefore, the n!
xn x
n 1 l0 1 so, by the Ratio Test, the series converges for all x and the radius of convergence is
R
.
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
n0 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
fn a
fx
x an
n!
n0
As with any convergent series, this means that f x is the limit of the sequence of partial
sums. In the case of the Taylor series, the partial sums are
n Tn x
i0 fa f i i! a x fa
1! a
x i a fa
2! x a 2 f n n! a x a n 5E12(pp 796805) 1/18/06 10:31 AM Page 799 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES y y=T£(x) T1 x y=T™(x)
(0, 1) y=T¡(x) 0 799 Notice that Tn is a polynomial of degree n called the nthdegree Taylor polynomial of f
at a. For instance, for the exponential function f x
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) ❙❙❙❙ x 1 x T2 x 1 x2
2! x T3 x 1 x2
2! x x3
3! The graphs of the exponential function and these three Taylor polynomials are drawn in
Figure 1.
In general, f x is the sum of its Taylor series if y=T£(x) fx
FIGURE 1
 As n increases, Tn x appears to approach
e x in Figure 1. This suggests that e x is equal to
the sum of its Taylor series. lim Tn x nl If we let
Rn x fx Tn x so that fx Tn x Rn x then Rn x is called the remainder of the Taylor series. If we can somehow show that
lim n l Rn x
0, then it follows that
lim Tn x nl lim f x Rn x nl fx lim Rn x nl fx We have therefore proved the following.
8 Theorem If f x
mial of f at a and Tn x Rn x , where Tn is the nthdegree Taylor polynolim Rn x 0 nl for x a
R, then f is equal to the sum of its Taylor series on the interval
xa
R. In trying to show that lim n l Rn x
following fact.
Taylor’s Inequality If f 0 for a speciﬁc function f , we usually use the n1 x
M for x
of the Taylor series satisﬁes the inequality
9 Rn x M
n 1! x a a n1 d, then the remainder Rn x for x To see why this is true for n 1, we assume that f x
fx
M , so for a x a d we have y x a f t dt y x a a d M . In particular, we have M dt An antiderivative of f is f , so by Part 2 of the Fundamental Theorem of Calculus, we
have
fx fa Mx a or fx fa Mx a 5E12(pp 796805) 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 n 1 is continuous on an interval I and
x I , then
1x
Rn x
y x t n f n 1 t 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
fn1z
Rn x
x an 1
n 1!
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 fx fx
But R1 x fa fx fx Mt a M
x
2 a M
x
2 a dt
x M a 2 2 2 a . So fax R1 x
A similar argument, using f x fa fax a fa x a fa fax T1 x y f t dt a 2 M , shows that
M
x
2 R1 x a 2 www.stewartcalculus.com
Click on Additional Topics and then on Formulas
for the Remainder Term in Taylor series. So M
x
2 R1 x a 2 Although we have assumed that x a, similar calculations show that this inequality is
also true for x 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 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 nl xn
n! for every real number x 0 This is true because we know from Example 1 that the series
and so its nth term approaches 0. x n n! converges for all x EXAMPLE 2 Prove that e x is equal to the sum of its Maclaurin series.
SOLUTION If f x x
d, then f
says that e x, then f n
x
ex 1 n1 x
e x for all n. If d is any positive number and
d
e . So Taylor’s Inequality, with a 0 and M e d,
ed Rn x n nl ed
n x n1 for x d e d works for every value of n. But, from Equa Notice that the same constant M
tion 10, we have
lim 1! 1! x n1 e d lim
nl x
n n1 1! 0 5E12(pp 796805) 1/18/06 10:32 AM Page 801 ❙❙❙❙ S ECTION 12.10 TAYLOR AND MACLAURIN SERIES 801 0 and therefore
It follows from the Squeeze Theorem that lim n l Rn x
lim n l Rn x
0 for all values of x. By Theorem 8, e x is equal to the sum of its
Maclaurin series, that is,
ex 11 n0  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:
1
n! e 12 n0 1
1! 1 1
2! 1
3! www.numbers.computation.free.fr e x at a EXAMPLE 3 Find the Taylor series for f x
SOLUTION We have f
(6), we get n 2 2 e and so, putting a
f n 2 x n! n0 2 2. 2 in the deﬁnition of a Taylor series n
n0 e2
x
n! 2 n Again it can be veriﬁed, as in Example 1, that the radius of convergence is R
0, so
Example 2 we can verify that lim n l Rn x
ex 13 n0 e2
x
n! 2 n . As in 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: fx sin x f0 0 fx cos x f0 1
0 fx sin x f0 f cos x f0 f x
4 x sin x f 4 0 1
0 Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as
follows:
f0
f0 2
f03
f0
x
x
x
1!
2!
3!
x x3
3! x5
5! x7
7! 1
n0 n x 2n 1
2n 1 ! 5E12(pp 796805) 802 ❙❙❙❙ 1/18/06 10:33 AM Page 802 CHAPTER 12 INFINITE SEQUENCES AND SERIES Since f
take M  Figure 2 shows the graph of sin x together
with its Taylor (or Maclaurin) polynomials
T1 x x x3
3! x x3
3! n1 x is sin x or cos x, we know that f
1 in Taylor’s Inequality: x 1 for all x. So we can x T3 x n1 T5 x x5
5! M Rn x 14 n 1! xn x 1 n1 n 1! By Equation 10 the right side of this inequality approaches 0 as n l , so
Rn x l 0 by the Squeeze Theorem. It follows that Rn x l 0 as n l , so sin x
is equal to the sum of its Maclaurin series by Theorem 8. Notice that, as n increases, Tn x becomes a
better approximation to sin x.
y We state the result of Example 4 for future reference. T¡
1 T∞ y=sin x sin x 15
0 x3
3! x x 1 n0 T£ x7
7! x 2n 1
2n 1 ! n 1 FIGURE 2 x5
5! 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
sin x
dx
1  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. 3x 2
3! d
dx x3
3! x 5x 4
5! x5
5! 7x 6
7! x7
7!
x2
2! 1 x4
4! x6
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 cos x 16 1 x2
2!
1 n0 x4
4!
n x 2n
2n ! x6
6!
for all x EXAMPLE 6 Find the Maclaurin series for the function f x 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 1
n0 EXAMPLE 7 Represent f x n x 2n
2n ! 1
n0 n x 2n 1
2n ! sin x as the sum of its Taylor series centered at 3. 5E12(pp 796805) 1/18/06 10:34 AM Page 803 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 803 SOLUTION Arranging our work in columns, we have fx
fx
 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 cos x 3 f s3
2 3 f x 1
2 1
2 and this pattern repeats indeﬁnitely. Therefore, the Taylor series at
f f 3
1! 3
s3
2 x sin x s3
2 3 f f y=sin x π
3 cos x 3 f fx f 0 sin x x
1 3
x 2 1! x
x x 3
3 1 3 3 3
3! 3
2 s3
2 2! 3 f 2 3
2! 3 is x 2 3! 3 The proof that this series represents sin x for all x is very similar to that in Example 4.
[Just replace x by x
3 in (14).] We can write the series in sigma notation if we
separate the terms that contain s3 : T£ sin x FIGURE 3 n0 1 ns3
x
2 2n ! 2n 1 3 n0 2 2n n 1! 2n 1 x 3 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
cn x a n
Theorem 5 asserts that, no matter how a power series representation f x
n
is obtained, it is always true that cn f a n!. 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 1
1 x ex
n0 xn 1 x2 x x3 1, 1 x3
3! , n0 xn
n! x
1! 1 x2
2! 1 n x 2n 1
2n 1 ! 1 sin x n x 2n
2n ! n0 cos x
n0 tan 1x 1
n0 n x 2n 1
2n 1 x x3
3! x2
2! 1
x x5
5!
x4
4! x3
3 x7
7!
x6
6! x5
5 ,
, x7
7 1, 1 5E12(pp 796805) 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 x
e x 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 e x dx as an inﬁnite series.
2
(b) Evaluate x01 e x dx correct to within an error of 0.001.
SOLUTION 2 (a) First we ﬁnd the Maclaurin series for f x
e x . Although it’s possible to use the
direct method, let’s ﬁnd it simply by replacing x with x 2 in the series for e x given in
the table of Maclaurin series. Thus, for all values of x,
x2
n! x2 e n0 n 1
n0 n x 2n
n! x2
1! 1 x4
2! x6
3! Now we integrate term by term: ye x2 y dx x2
1! 1 C x x4
2! x3
3 1! x6
3! 1 x5
5 2! n x 2n
n! dx x7
7 3! 1 This series converges for all x because the original series for e
(b) The Evaluation Theorem gives  We can take C
in part (a). 0 in the antiderivative y 1 0 e x2 dx x3
3 1! x x5
5 2! 1 1
3 1
10 1
42 1
3 1
10 1
42 1
216 x 2n 1
2n 1 n! converges for all x. x9
9 4! 1 0 1
216 1 x7
7 3! x2 n 0.7475 The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than
1
11 5! 1
1320 0.001 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 1
x2 x . 5E12(pp 796805) 1/18/06 10:35 AM Page 805 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 805 SOLUTION Using the Maclaurin series for e x, we have  Some computer algebra systems compute
limits in this way. lim xl0 ex 1
x2 1 x x
1! x2
2! x2
2! x3
3! x4
4! lim x3
3!
x2 xl0 lim xl0 x x2 xl0 lim 1 1
2 x2
4! x
3! x3
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 x2
2! x
1! 1 x3
3! x x3
3! We multiply these expressions, collecting like terms just as for polynomials:
1
6
1
6 x3
x3 x2 1
2
1
6 x3
x3 x
Thus x2 x 1 x2 1
3 x3 x
x 1
2 e x sin x x2 x 1
6
1
6 1
3 x4
x4 x3 (b) Using the Maclaurin series in the table, we have tan x sin x
cos x x
1 x3
3!
x2
2! x5
5!
x4
4! 5E12(pp 806815) ❙❙❙❙ 806 1/18/06 10:18 AM Page 806 CHAPTER 12 INFINITE SEQUENCES AND SERIES We use a procedure like long division:
x
x2 Thus 1
24 x
x x4 tan x 1
3 x x5 x3
x3 2
15
1
120
1
24 x3
x3 1
30
1
6 x5
x5 2
15 1
2 x3 1
6
1
2
1
3
1
3 1 1
3 x5 2
15 x3 x5
x5 x5 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 x
cn x n and
n
tx
bn x converge for x
R and the series are multiplied as if they were polynomials, then the resulting series also converges for x
R and represents f x t x . For
division we require b0 0; the resulting series converges for sufﬁciently small x .  12.10
1. If f x Exercises
n0 5 n for all x, write a formula for b 8. bn x 11–18  Find the Taylor series for f x centered at the given value
of a. [Assume that f has a power series expansion. Do not show
that Rn x l 0.] 2. (a) The graph of f is shown. Explain why the series 1.6 0.8 x 1 0.4 x 1 2 0.1 x 1 3 11. f x 1 x 2, a x 12. f x 0 cos x,
sin x, a 1 sx, a 18. f x 2 ■ (b) Explain why the series
2.8 0.5 x 2 1.5 x 2 0.1 x 2 a ■ ■ 2
9
1
■ ■ ■ ■ ■ ■ 5. f x 1 7. f x 21. Prove that the series obtained in Exercise 9 represents sinh x for all x.
22. Prove that the series obtained in Exercise 10 represents cosh x for all x. sin 2 x 6. f x ln 1 8. f x xe x 23. f x cos x 24. f x e 9. f x 10. f x cosh x 25. f x x tan 1x 26. f x sin x 4 ■ ■ 27. f x x 2e 28. f x x cos 2 x 3 e5x
sinh x x ■ ■ ■ ■ for all x. 4. f x cos x ■ x, 2 20. Prove that the series obtained in Exercise 16 represents sin x  Find the Maclaurin series for f x using the deﬁnition of
a Maclaurin series. [Assume that f has a power series expansion.
Do not show that Rn x l 0.] Also ﬁnd the associated radius of
convergence. ■ a all x. 3 3–10 ■ ■ 1
3 ■ 19. Prove that the series obtained in Exercise 3 represents cos x for 2 is not the Taylor series of f centered at 2. 3. f x 2 a 17. f x x 1 ln x, 16. f x 1 a 15. f x f a e x, 14. f x y x, 13. f x is not the Taylor series of f centered at 1. 3 ■ 23–32  Use a Maclaurin series derived in this section to obtain
the Maclaurin series for the given function. x ■ ■ ■ x x2 5E12(pp 806815) 1/18/06 10:18 AM Page 807 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES sin 2x 29. f x 1 cos 2 x .] 49. lim
■ sin x
x
1 31. f x x
32. f x if x 0 if x if x ■ 0 ■ ■ ■ ■ ■ 34. f x 2 ■ ■ x2 e 36. f x
■ ■ 2 ■ ■ e 53. y 0.2 ■ ■ correct to ﬁve 55–60 ■ ■ 3 40. dx 1 42. 1 dx ■ ■ ■ ■ ■ y ■ ■ ■ 57. ■  Use series to approximate the deﬁnite integral to within
the indicated accuracy. y 44. y 45.
46. x cos x 3 d x 0.2 ln 2
■ 0.5 0 2 xe ■ x2 ■  47. lim xl0 x f sin x 3 d x ■ error 10 ■ ■ ■ n0 ■ n ■ ■ 2n 2n ! n 58. 1! n0 3
5n n! 81
4! ln 2
2! ■ 6 2n 2 3 ln 2
3! ■ ■ ■ ■ ■ ■ error M for x 2, that is, prove that if
d, then a M
x
6 0.001) ■ ■ R2 x ) ■ a 3 for x a d 62. (a) Show that the function deﬁned by ■ ■ ■ ■ fx ■ Use series to evaluate the limit.
tan 1x
x3 x 8 ( dx (ﬁve decimal places) ( x3 s1 0 47–49 ■ 1 2n 1 27
3! 9
2! ■ (three decimal places) x3 dx 0.1 ■ 1 tan 0 y ■ x 61. Prove Taylor’s Inequality for n 1 0 y 4 1
2n 60. 1 ■ 43–46 43. ■ 56. 2n 1 59. 3 ■ ■ x 4n
n! n n0 sin x
dx
x
ex 1
dx
x y ■ e x ln 1 Find the sum of the series. 55. Evaluate the indeﬁnite integral as an inﬁnite series. y sx ■ sec x 54. y ■  n0 41. ■ x 52. y cos x n y x cos x ■ ■ ﬁve decimal places. 39. tan x
x3 x
sin x x
■ x2 51. y 38. Use the Maclaurin series for sin x to compute sin 3 correct to 3 ■  Use multiplication or division of power series to ﬁnd the
ﬁrst three nonzero terms in the Maclaurin series for each function. decimal places.  ■ 51–54 cos x ■ ■ 37. Use the Maclaurin series for e x to calculate e 39–42 ■ ■ ■
■ ■ We found this limit in Example 4 in Section 7.7 using l’Hospital’s Rule three times. Which method do you prefer? x cos 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 35. f x ■ 50. Use the series in Example 10(b) to evaluate if x s1 x3 5 lim sin x
x3 ■ 33. f x x ■ 1
6 x 807 0 1
6 ■ sin x xl0 cos 2x 30. f x ■ 1
2 [Hint: Use sin 2x ❙❙❙❙ 1
48. lim
xl0 1 cos x
x
ex ; e
0 1 x2 if x
if x 0
0 is not equal to its Maclaurin series.
(b) Graph the function in part (a) and comment on its behavior
near the origin. 5E12(pp 806815) 808 ❙❙❙❙ 1/18/06 10:18 AM Page 808 CHAPTER 12 INFINITE SEQUENCES AND SERIES LABORATORY PROJECT
C AS An Elusive Limit
This project deals with the function
fx sin tan x
arcsin arctan x tan sin x
arctan arcsin x 1. Use your computer algebra system to evaluate f x 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 x 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 x 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 x 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 b k ak ka k 1b kk 1 a k 2b 2 2!
kk 1k kk 2 k 1k
3!
n 2 1 a k nb n n!
kab k 1 a k 3b 3 bk The traditional notation for the binomial coefﬁcients is
k
0 1 k
n kk 1k 2 k n n! 1 n which enables us to write the Binomial Theorem in the abbreviated form
k a b k
n0 k knn
ab
n 1, 2, . . . , k 5E12(pp 806815) 1/18/06 10:19 AM Page 809 SECTION 12.11 THE BINOMIAL SERIES In particular, if we put a 1 and b 809 x, we get
k 1 1 ❙❙❙❙ kn
x
n k x n0 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 x k is no longer a ﬁnite sum; it becomes an inﬁnite
series. To ﬁnd this series we compute the Maclaurin series of 1 x k in the usual way:
k fx 1 x f0 fx k1 x fx kk 11 x f x kk 1k 21 f0 x kk 1 k n k f0 kk 1 f0
.
.
.
n
f0 kk 1k 2 kk 1 k k1 k2 k3 x .
.
.
f 1 n 11 x Therefore, the Maclaurin series of f x
f n 0 n! n0 kn 1 x k is 1
kk xn n 1 k n 1 n! n0 xn This series is called the binomial series. If its nth term is a n , then
an 1
an kk k
n 1 n
1 k
n n xn n 1k
1!
k
n
1
n 1
x
1 1 kk xlx n!
k 1 n 1 xn as n l Thus, by the Ratio Test, the binomial series converges if x
1 and diverges if x
1
.
The following theorem states that 1 x k is equal to the sum of its Maclaurin series.
It is possible to prove this by showing that the remainder term Rn x approaches 0, but that
turns out to be quite difﬁcult. The proof outlined in Exercise 19 is much easier.
2 The Binomial Series If k is any real number and x 1 x k 1 k
n kk 1
2! x2 kk 1k
3! 2 x3 kn
x
n n0 where kk kx 1, then 1 k
n! n 1 n 1 and k
0 1 5E12(pp 806815) 810 ❙❙❙❙ 1/18/06 10:19 AM Page 810 CHAPTER 12 INFINITE SEQUENCES AND SERIES 1, the question of whether
Although the binomial series always converges when x
or not it converges at the endpoints, 1, depends on the value of k. It turns out that the
series converges at 1 if 1 k 0 and at both endpoints if k 0. Notice that if k is a
k
k
positive integer and n k, then the expression for ( n ) contains a factor k k , so ( n ) 0
for n 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 EXAMPLE 1 Expand 1 x 2 as a power series. SOLUTION We use the binomial series with k 2
n 2 3 2. The binomial coefﬁcient is 4 2 n 1 n!
1 n2 3 4 nn 1 1nn n!
and so, when x 1 1,
1 1 x 1 2 2n
x
n 2 x n0 1nn 1 xn 1 3x 2 2x 4x3 n0 1 EXAMPLE 2 Find the Maclaurin series for the function f x s4 convergence.
SOLUTION As given, f x is not quite of the form 1 1
s4 1
x x
4
1
2 Using the binomial series with k
1
s4 x 1
2 1 1
2 x
4 1 12 1
2 1
2 1
2
n0 x
4 2 1
2 n!
1
2 1 1
x
8 132
x
2!8 2 x
4 2! ( x
4 1353
x
3!8 3 12 x
4 1 x 4, we have n ( 1 )( 3 )
2
2 ( 1)( 3)( 5)
2
2
2 1
2 and with x replaced by x
4 n 1 and its radius of x k so we rewrite it as follows: 1 41 x n 2 ( 1)( 3)( 5)
2
2
2 x
4 3!
1) x
4 3 n 135 2n
n!8 n 1 xn 5E12(pp 806815) 1/18/06 10:19 AM Page 811 SECTION 12.11 THE BINOMIAL SERIES x4 We know from (2) that this series converges when
radius of convergence is R 4.
y y=
1 1, that is, x 1
œ4x
œ„„„„ ❙❙❙❙ 811 4, so the 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.
0 _4 x 4 FIGURE 1  12.11
1–8 Exercises
x 1 x 2 as a power series.
(b) Use part (a) to ﬁnd the sum of the series 15. (a) Expand f x Use the binomial series to expand the function as a power
series. State the radius of convergence.
 1. s1 1 3. 2 x 4
5. s1 7. 2. x ■ ; 9–10 x n1 x x 2 1 x 3 as a power series.
(b) Use part (a) to ﬁnd the sum of the series 16. (a) Expand f x 23 1
5
s32 x
x2
8.
s2 x x2 ■ x 4 6. 8x
x s4 1 4. 1 3 ■ ■ ■ ■ ■ ■ ■ n1 ■ ■ ■ ■ 34 ■ 3
10. s1 ■ ■ ■ ■ ■ fx
1 s1 x 3.
(b) Use part (a) to evaluate f ■ (a) Let t x
■ ■ 11. (a) Use the binomial series to expand 1 s1 x 2.
(b) Use part (a) to ﬁnd the Maclaurin series for sin 1x. 12. (a) Use the binomial series to expand 1 s1 x 2.
(b) Use part (a) to ﬁnd the Maclaurin series for sinh 1x. 13. (a) Expand s1 n0 4
14. (a) Expand 1 s1 x as a power series.
4
(b) Use part (a) to estimate 1 s1.1 correct to three decimal
places. 0. 9 0. k
( n )x n. Differentiate this series to show that kt x
1x tx ■ (b) Let h x
1x
(c) Deduce that t x 1 x 1 k t x and show that h x
1 x k. 0. 20. In Exercise 53 in Section 11.2 it was shown that the length of the ellipse x a sin , y 3 x as a power series.
3
(b) Use part (a) to estimate s1.01 correct to four decimal
places. 10 19. Use the following steps to prove (2). 4x ■ fx
s1 x 2.
(b) Use part (a) to evaluate f 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.
2x n2
2n 17. (a) Use the binomial series to ﬁnd the Maclaurin series of  9. 1 n
2n 1 L 4a y 0 2 a cos , where a
s1 b 0, is e 2 sin2 d where e sa 2 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. 5E12(pp 806815) 812 ❙❙❙❙ 1/18/06 10:19 AM Page 812 CHAPTER 12 INFINITE SEQUENCES AND SERIES WRITING PROJECT
H ow Newton Discovered the Binomial Series
The Binomial Theorem, which gives the expansion of a b k, 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 b k 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 x 2 n 2 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 x 2 n 2 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 x 2 n 2. 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: SpringerVerlag, 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 x is equal to the sum of its Taylor series at a:
f fx
n0 n n! a x a n 5E12(pp 806815) 1/18/06 10:20 AM Page 813 ❙❙❙❙ S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS 813 In Section 12.10 we introduced the notation Tn x for the nth partial sum of this series and
called it the n thdegree Taylor polynomial of f at a. Thus
n Tn x
i0 fa f i i! a x fa
1! a i x a fa
2! x a f 2 n a n! Since f is the sum of its Taylor series, we know that Tn x l f x as n l
be used as an approximation to f : f x
Tn x .
Notice that the ﬁrstdegree Taylor polynomial x a n and so Tn can y T1 x 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 T2 x
T4 x
T6 x
T8 x
T10 x 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 fa fax a 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 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, 1 ; this tangent line is the best linear approximation to e x near 0, 1 . The graph
of T2 is the parabola y 1 x x 2 2, and the graph of T3 is the cubic curve
y 1 x x 2 2 x 3 6, 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 Tn x 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 Tn x 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:
Rn x fx Tn x There are three possible methods for estimating the size of the error:
1. If a graphing device is available, we can use it to graph Rn x and thereby estimate 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
fn1 x
M , then
Rn x M
n 1! x a n1 EXAMPLE 1
3
(a) Approximate the function f x
sx by a Taylor polynomial of degree 2 at a
(b) How accurate is this approximation when 7 x 9? 8. 5E12(pp 806815) 814 ❙❙❙❙ 1/18/06 10:20 AM Page 814 CHAPTER 12 INFINITE SEQUENCES AND SERIES S OLUTION (a) 1
3 fx 2
9
10
27 x f8 x 53 2 f8 23 x fx
f x1 3 3
sx fx 1
12
1
144 f8 83 x Thus, the seconddegree Taylor polynomial is
T2 x f8
1! f8
2 1
12 x x 1
288 8 f8
2! 8
x 8 x 8 2 2 The desired approximation is
3
sx T2 x 1
12 2 x 1
288 8 x 8 2 (b) The Taylor series is not alternating when x 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
x
3! R2 x
where f x M . Because x
f x 7, we have x 8 3
10
27 1
x 10
27 83 Therefore, we can take M 0.0021. Also 7
x8
1. Then Taylor’s Inequality gives 2.5 T™ R2 x 8 0.0021
3! x 3 7 8 3 and so
1
78 3 0.0021 9, so 1 0.0021
6 13 x 8 1 and 0.0004 #„
y= œx Thus, if 7 x 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 T2 x are very close to each other when x is near 8. Figure
3 shows the graph of R2 x computed from the expression FIGURE 2
0.0003 R2 x
y=R™(x) T2 x We see from the graph that
R2 x 7 3
sx 0.0003 9
0 FIGURE 3 when 7 x 9. Thus, the error estimate from graphical methods is slightly better than
the error estimate from Taylor’s Inequality in this case. 5E12(pp 806815) 1/18/06 10:20 AM Page 815 S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 815 EXAMPLE 2 (a) What is the maximum error possible in using the approximation
sin x x3
3! x x5
5! when 0.3 x 0.3? Use this approximation to ﬁnd sin 12 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
sin x x3
3! x x5
5! x7
7! is alternating for all nonzero values of x, and the successive terms decrease in size
because x
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
x7
7!
If 0.3 x 0.3, then x x7
5040 0.3, so the error is smaller than
0.3 7
5040 4.3 10 8 To ﬁnd sin 12 we ﬁrst convert to radian measure.
sin 12 sin 12
180 sin
3 15 15 15
5 1
3! 15 1
5! 0.20791169
Thus, correct to six decimal places, sin 12
0.207912.
(b) The error will be smaller than 0.00005 if
x7
5040 0.00005 Solving this inequality for x, we get
x 7 0.252 or x 0.252 17 0.821 So the given approximation is accurate to within 0.00005 when x
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
have f 7 x
1 and so
1
R6 x
x7
7! 0.82.
7 x cos x, we So we get the same estimates as with the Alternating Series Estimation Theorem. 5E12(pp 816824) 816 ❙❙❙❙ 1/18/06 10:22 AM Page 816 CHAPTER 12 INFINITE SEQUENCES AND SERIES 4.3 What about graphical methods? Figure 4 shows the graph of 10–* R6 x (x sin x y= Rß(x) _0.3 0.3
0 FIGURE 4
0.00006
y=0.00005 1
6 x3 T1 x
1
0 FIGURE 5 x5) and we see from it that R6 x
4.3 10 8 when x
0.3. This is the same estimate
that we obtained in Example 2. For part (b) we want R6 x
0.00005, so we graph both
y
R6 x and y 0.00005 in Figure 5. By placing the cursor on the right intersection
point we ﬁnd that the inequality is satisﬁed when x
0.82. Again this is the same estimate that we obtained in the solution to Example 2.
If we had been asked to approximate sin 72 instead of sin 12 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 72 is close to
60 (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 y= Rß(x) _1 1
120 T5 x x T3 x
x3
3! x5
5! x x3
3! T7 x x x x3
3! x5
5! x7
7! to the sine curve. You can see that as n increases, Tn x is a good approximation to sin x on
a larger and larger interval.
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 v 2 c 2 where m0 is the mass of the object when at rest and c is the speed of light. The kinetic 5E12(pp 816824) 1/18/06 10:22 AM Page 817 S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 817 energy of the object is the difference between its total energy and its energy at rest:
mc 2 K 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
100 m s.
SOLUTION (a) Using the expressions given for K and m, we get
mc 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 12 x 1 1
2 1 1 and 1 c2 1
2 ( 1 )( 3 ) x 2
2
2 x 3
8 1 ( 1 )( 3 )( 5) x 3
2
2
2 2! x 5
16 2 1 v2
2 c2 m0 c 2 K x 1 2 is most easily computed as a
1 because v c.) Therefore, we have 1 v2
2 c2 x 3! 3 3 v4
8 c4 3 v4
8 c4 5 v6
16 c 6 1 5 v6
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 12 x m0 c 2 0 v2 m0 c 2 v2 c2 s1 With x
v 2 c 2, the Maclaurin series for 1
1
binomial series with k
2 . (Notice that x K K=mc@m¸c@ m0 c 2 m0 c 2 m0 c 2 K √ m0 c 2 K FIGURE 7 (b) If x
fx 1 v2
2 c2 1
2 m0 v 2 v 2 c 2, f x m0 c 2 1 x 1 2 1 , and M is a number such that
M , then we can use Taylor’s Inequality to write
M2
x
2! R1 x
3
4 We have f x m0 c 2 1 x and we are given that v 3m0 c 2
4 1 v2 c2 fx
3 1
2 52 41 3m0 c 2
100 2 c 2 100 m s, so
M 52 10 8 m s, R1 x Thus, with c 52 41 3m0 c 2
100 2 c 2 52 100 4
c4 4.17 10 10 m0 So when v
100 m s, the magnitude of the error in using the Newtonian expression
for kinetic energy is at most 4.2 10 10 m0. 5E12(pp 816824) 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: AddisonWesley, 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 R V ˙ ¨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
n1 n2 o 1 i 1
R n2si n1so i o where n1 and n2 are indexes of refraction and o , i , so , and si are the distances indicated in
Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have cos o cos sR 2 so R 2 2R so R cos i 2  Here we use the identity sR 2 si R 2 2R si R cos Because Equation 1 is cumbersome to work with, Gauss, in 1841, simpliﬁed it by using
1 for small values of . (This amounts to using the
the linear approximation cos
Taylor polynomial of degree 1.) Then Equation 1 becomes the following simpler equation
[as you are asked to show in Exercise 32(a)]:
n1
so 3 n2
si n2 n1
R The resulting optical theory is known as Gaussian optics, or ﬁrstorder 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 n1
so n2
si n2 n1
R h2 n1
2so 1
so 1
R 2 n2
2si 1
R 1
si 2 The resulting optical theory is known as thirdorder 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. 5E12(pp 816824) 1/18/06 10:23 AM Page 819 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS  12.12 20. f x a ln x, 4. f x x e, 5. f x sin x, a 6. f x cos x, a 7. f x arcsin x, 8. f x ln x
,
x 10. f x
■ a 1, a 2x xe n 2, 2 a 1, a , 0, s3 x, a ■ ■ n ■ 27. sin x 12. f x tan x, n ■ ■ n n ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 4, n 2, 4 a 1, n 2, 0.9 x ■ a 1, n 3, 0.8 x 15. f x x 23 16. f x cos x, a 17. f x tan x, a 18. f x ln 1 2x , 19. f x x2 e, , a 3, n
0, 0, n a
n ■ ■ ■ 0.01) error
x4
24 ( ■ error ■ x t 1.2 x, x2
2 ( 0.005) ■ ■ ■ ■ ■ ■ of a conducting wire is the reciprocal of the
conductivity and is measured in units of ohmmeters ( m).
The resistivity of a given metal depends on the temperature
according to the equation ■ 1.1 14. f x x do you 30. The resistivity
■ a 2 ■ at a given instant. Using a seconddegree 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?  sx, ■ 29. A car is moving with speed 20 m s and acceleration 2 m s2 (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 x
Tn x when x lies in the given interval.
; (c) Check your result in part (b) by graphing Rn x .
13. f x 1 ■ 1, 3, 5, 7, 9
■ x3
6 28. cos x 2 2, 4, 6, 8 ■ x 3  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. sec x, ■  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. 11–12 11. f x 1 ■ ; 27–28
4 3 1, x ■ 1 the Maclaurin series for e x that should be used to estimate e 0.1
to within 0.00001. 3 n 2 1 ■ x 25. Use Taylor’s Inequality to determine the number of terms of 3 n n 5, ■ 1 26. How many terms of the Maclaurin series for ln 1 n 0, n 0,
■ 4, 1.5 need to use to estimate ln 1.4 to within 0.001? 3, a n x correct to ﬁve decimal places. 3
6, ■ 0, 0.5 24. Use the information from Exercise 16 to estimate cos 69 4 n ■ 3, n correct to ﬁve decimal places.  3. f x 13–22 sinh 2 x, ■ 1, a 23. Use the information from Exercise 5 to estimate sin 35 Find the Taylor polynomial Tn x for the function f at the
number a. Graph f and Tn on the same screen. ■ a ■ 1x
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 x . CAS x sin x, 22. f x ; 2. (a) Find the Taylor polynomials up to degree 3 for f x ■ x ln x, 21. f x fx
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 x . 9. f x 819 Exercises ; 1. (a) Find the Taylor polynomials up to degree 6 for ; 3–10 ❙❙❙❙ 4,
3, 1, n
3, 0 4.2 0 0 x
x 3,
x 2 3 6 0.5
0.1 x ;
1.5 20 e t 20 where t is the temperature in C. There are tables that list the
values of (called the temperature coefﬁcient) and 20 (the
resistivity at 20 C) 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 t by
its ﬁrst or seconddegree Taylor polynomial at t 20.
(a) Find expressions for these linear and quadratic
approximations.
0.0039 C and
(b) For copper, the tables give
1.7 10 8 m. Graph the resistivity of copper
20
and the linear and quadratic approximations for
250 C t 1000 C. 5E12(pp 816824) ❙❙❙❙ 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 q and
are located at a distance d from each other, then the electric
ﬁeld E at the point P in the ﬁgure is
E q q
D2 D d 2 By expanding this expression for E as a series in powers of
d D, show that E is approximately proportional to 1 D 3 when
P is far away from the dipole.
q _q P
d D 32. (a) Derive Equation 3 for Gaussian optics from Equation 1 by approximating cos in Equation 2 by its ﬁrstdegree Taylor
polynomial.
(b) Show that if cos is replaced by its thirddegree Taylor
polynomial in Equation 2, then Equation 1 becomes Equation 4 for thirdorder optics. [Hint: Use the ﬁrst two terms
in the binomial series for o 1 and i 1. Also, use
sin .]
33. If a water wave with length L moves with velocity v across a body of water with depth d, as in the ﬁgure, then
tL
2d
tanh
2
L v2 (a) If the water is deep, show that v stL 2 .
(b) If the water is shallow, use the Maclaurin series for tanh to
show that v 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 10d, then the estimate v 2 td is accurate to within
0.014tL. 0 L
(1
t 1
4 k2) T L4
t4 2 3k 2
4k 2 (c) Use the inequalities in part (b) to estimate the period of a
pendulum with L 1 meter and 0 10 . How does it compare with the estimate T 2 sL t ? What if 0 42 ?
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 sec L R R (b) Use a Taylor polynomial to show that
L2
2R C 5L 4
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 mating a root r of the equation f x
0, and from an initial
approximation x 1 we obtained successive approximations x 2 ,
x 3 , . . . , where with the vertical is T L
t 4 y xn 2 0 s1 dx
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
2 2 37. In Section 4.9 we considered Newton’s method for approxi 34. The period of a pendulum with length L that makes a maxi T (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 36. Show that Tn and f have the same derivatives at a up to order n. L
d mum angle If 0 is not too large, the approximation T 2 sL t,
obtained by using only the ﬁrst term in the series, is often
used. A better approximation is obtained by using two
terms:
L
T2
(1 1 k 2 )
4
t L
t 1 12 2
k
22 12 3 2 4
k
2 242 12 3 25 2 6
k
2 2426 2 xn 1 f xn
f xn Use Taylor’s Inequality with n 1, a x n, and x r to show
that if f x exists on an interval I containing r, x n , and x n 1,
and f x
M, f x
K for all x I , then
xn 1 r M
xn
2K r 2 [This means that if x n is accurate to d decimal places, then x n 1
is accurate to about 2d decimal places. More precisely, if the
error at stage n is at most 10 m, then the error at stage n 1 is
at most M 2K 10 2 m.] 5E12(pp 816824) 1/18/06 10:24 AM Page 821 A PPLIED PROJECT RADIATION FROM THE STARS ❙❙❙❙ 821 A PPLIED 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 RayleighJeans Law expresses the energy density of
blackbody radiation of wavelength as
8 kT f 4 where is measured in meters, T is the temperature in kelvins (K), and k is Boltzmann’s constant. The RayleighJeans Law agrees with experimental measurements for long wavelengths but
disagrees drastically for short wavelengths. [The law predicts that f
l as l 0 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:
8 hc f
where e hc 5 kT 1 is measured in meters, T is the temperature (in kelvins), and
h Planck’s constant 6.6262 c speed of light k Boltzmann’s constant 10 34 Js 8 2.997925 10 m s 1.3807 10 23 JK 1. Use l’Hospital’s Rule to show that lim f
l0 0 and lim f
l 0 for Planck’s Law. So this law models blackbody radiation better than the RayleighJeans
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 RayleighJeans 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
10 6 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. 5E12(pp 816824) ❙❙❙❙ 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 an 3 mean?
(d) What does n 1 an 3 mean? 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 x 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 ?
(b) How do you integrate f ? What is the radius of convergence
of the series for x f x d x? convergent? What is its sum?
(b) What is a pseries? 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 an? What is lim n l sn? 10. (a) Write an expression for the nthdegree 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 x 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) 1 1 x
(b) e x
(c) sin x
(d) cos x
(e) tan 1x (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. The series
3. If lim n l a n 0, then
n1 n sin 1 a n is convergent. 1 4. If cn6 n is convergent, then cn 5. If cn6 n is convergent, then cn 6 n is convergent. 6. If cn x n diverges when x 6, then it diverges when x 7. The Ratio Test can be used to determine whether 1 n3 converges.
8. The Ratio Test can be used to determine whether n0 n 1
e 1 a n is divergent, then f0 2x
2. x2 bn diverges, then 1
3 0. a n is divergent. x3 converges for all x, then 14. If a n and bn are divergent, then a n 10. 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 0 for all n, then a n is 1 n!
17. If a n bn and n 1, then lim n l convergent. converges.
an 1
n! 10. 13. If f x 2 n is convergent. 9. If 0 ■ 12. If L. x k. What is the radius of convergence of this series? 11. If is convergent. L, then lim n l a 2 n 12. Write the binomial series expansion of 1 TRUEFALSE 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. If lim n l a n ■ a n diverges. 0 and 18. If a n 0 and lim n l a n converges, then
an 1 an 1 n a n converges.
1, then lim n l a n 0. 5E12(pp 816824) 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. n3
2n 3 2
1 1. a n 2. a n 4. a n 1 n 5. a n
7. 3n ■ ■ cos n 6. a n 2 n sin n
n2 1 1 27–31 2 ln n
sn n1 4n 8. ■ ■ ■ ■ ■ n 10
■ 2 1
n0 n! 31. 1
■ ■ 9. A sequence is deﬁned recursively by the equations a 1 a n 1 1 a n 4 . Show that a n is increasing and a n
3
all n. Deduce that a n is convergent and ﬁnd its limit. 1,
2 for n1 n 3 12. 1 n1 n2
n3 14. n1 n
5n 16. n2 1
n s ln n 18. n1 cos 3n
1
1.2 13. 15. 17. n1 2n n1 sn 22.
■ n2n
1 2n2 ■ 1 n1
■ ■ n ■ 1
2 1 ■ 13 n1 5n 39. Prove that if the series sn n1 1 ■ ■ 24. 1 3n n ■ n1
■ ■ ■ ■ ■ ■ ■ ■ n1 ln x n converge? 1 correct to four n1 1 . Estimate the error involved in this
nn
is convergent.
2n !
0. n1 an is absolutely convergent, then the series 1 ■ ■ 1
n n 40–43  Find the radius of convergence and interval of convergence of the series. 41. 2n x 2 n
n 2! 43. 40. 1 nsn
ln n
■ xn
n 25 n n1 3 1 42.
■ ■ ■ an is also absolutely convergent. ■ n1 n1 1 26. 1
■ 1n
n5 n ■ ■ x 2 for all x. n1 n1 1nn
22n ■ n1
■ n1 ■ 1 n1 ■ ■ nn
(b) Deduce that lim
nl
2n !  Determine whether the series is conditionally convergent,
absolutely convergent, or divergent. 25. ■ 38. (a) Show that the series n 23–26 23. e4
4! n n1
■ ■ the series n 1 2
approximation. 1 n1 sn e3
3! 37. Use the sum of the ﬁrst eight terms to approximate the sum of n
3n 21. 1 e2
2! e 1 5 nn!
5 2n
n 9n 2n ! 35. Find the sum of the series 1 2 n 1 n 6 and estimate
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. n1 n n1 20. ln n 34. For what values of x does the series n sn 3 2n 36. (a) Find the partial sum s5 of the series n1 135 19. 1 3 tan 1n 1 decimal places. 1
1 3 n 33. Show that cosh x n n 11. 1 nn 32. Express the repeating decimal 4.17326326326 . . . as a fraction. Determine whether the series is convergent or divergent.  n1 ■ 0 and use a graph to ﬁnd the smallest value of N that corresponds to
0.1 in the precise deﬁnition of a limit. 11–22 1 28. 5n
tan ■ 4
; 10. Show that lim n l n e 2n 1 n1 30. ■ Find the sum of the series.  29. n3 3. a n 823 ■ 27. 9n 1
10 n ❙❙❙❙ ■ ■ n ■ x
n1 n0
■ ■ ■ ■ ■ 2
n4 n n 2n x
sn 3n
3 ■ ■ ■ ■ 5E12(pp 816824) 824 ❙❙❙❙ 1/18/06 10:26 AM Page 824 CHAPTER 12 INFINITE SEQUENCES AND SERIES 44. Find the radius of convergence of the series n1 57. f x
58. f x 2n ! n
x
n! 2 s x,
sec x, ■ 45. Find the Taylor series of f x sin x at a 6. 46. Find the Taylor series of f x cos x at a ■ a ■ 1,
a n
0, n ■ ■ xl0 47–54  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 tan 1x. 1 48. f x
50. f x 49. f x ln 1 51. f x
53. f x 4
1 s16 ■ ■ ■ 55. Evaluate y ■ x
■ ■ 1 ■ ■ ■ 3x ; 5 ■ ■ ■ x 4 d x correct to two decimal places.
57–58 6 ■ ■ ■ ■ ■ sin x x
x3 mtR2
Rh F x2 ex
d x as an inﬁnite series.
x 1
56. Use series to approximate x0 s1 x ■ above the surface of the Earth is 10 x 54. f x sin x 4 x 1 0 1.1 60. The force due to gravity on an object with mass m at a height h xe 2 x 52. f x x tan 2, x 59. Use series to evaluate the following limit. lim x2 0.9 ■ 3. 47. f x 3, 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 h R.
(b) Observe that if we approximate F by the ﬁrst term in the
series, we get the expression F m t 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 m t is accurate to within 1%.
(Use R 6400 km.)
cn x n for all x.
(a) If f is an odd function, show that 61. Suppose that f x n0 c0  (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 x
Tn x when x lies in the given interval.
; (d) Check your result in part (c) by graphing Rn x . 2 c2 c4 0 (b) If f is an even function, show that
c1
62. If f x 2 c3 e x , show that f c5
2n 0 0
2n !
.
n! 5E12(pp 825827) 1/18/06 10:27 AM Page 825 PROBLEMS
PLUS sin x 3 , ﬁnd f 1. If f x 15 0. 2. A function f is deﬁned by fx lim nl x 2n
x 2n 1
1 Where is f continuous?
1 cot 1 x
2 3. (a) Show that tan 2 x 2 cot x. (b) Find the sum of the series
n1 P¢ 4 4. Let Pn be a sequence of points determined as in the ﬁgure. Thus AP1 P£
2 8 P∞
FIGURE FOR PROBLEM 4 1
x
tan n
2n
2 P™
1
A 1 P¡ Pn Pn 1 2 n 1, and angle APn Pn 1 is a right angle. Find lim n l 1,
Pn APn 1 . 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 as n l .
(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 1 1
2 1
3 1
4 1
6 1
8 1
9 1
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 1,
arctan x if the left side lies between
(b) Show that arctan y
2 and 120 arctan 119 arctan x
1 y
xy 2. 1
arctan 239 4 (c) Deduce the following formula of John Machin (1680–1751):
4 arctan 1
5 1
arctan 239 4
825 5E12(pp 825827) 1/18/06 10:27 AM Page 826 (d) Use the Maclaurin series for arctan to show that
1 0.197395560 0.004184075 0.197395562 arctan 5 arctan 239 (e) Show that
1 0.004184077 (f) Deduce that, correct to seven decimal places,
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 Brent Salamin, 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
arccot n 2 n 1 n0 9. Find the interval of convergence of
10. If a 0 a1 0, show that a2 ak lim (a0 sn a1 sn nl n 3x n and ﬁnd its sum. n1 a2 sn 1 ak sn 2 k) 0 If you don’t see how to prove this, try the problemsolving 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. 1
.
n2 ln 1 11. Find the sum of the series
n2 12. Suppose you have a large supply of books, all the same size, and you stack them at the edge
1
16
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 onesixth of its length beyond the fourth, and so on. (Try it yourself
with a deck of cards.) Consider centers of mass.
u 14. If p v3 w3 x3
3! x6
6! x9
9! x x4
4! x7
7! x 10
10! w Show that u 3 1 v 13. Let x2
2! 3u vw x8
8! 1. 1, evaluate the expression
1
1 826 x5
5! 1
2p
1
2p 1
3p
1
3p 1
4p
1
4p 5E12(pp 825827) 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 nl An
A 2 s3 16. A sequence a n is deﬁned recursively by the equations a0 a1 1 Find the sum of the series FIGURE FOR PROBLEM 15 nn
n0 1 an n 1n 2 an n 1 3 an 2 an. x 17. Taking the value of x at 0 to be 1 and integrating a series termbyterm, show that y 1 0 P¡ P∞ P™
P˜ Pˆ Pß
P¡¸ 1n
nn x x dx
n1 1 18. Starting with the vertices P1 0, 1 , P2 1, 1 , P3 1, 0 , P4 0, 0 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 x n 1 x n 2 x n 3 2 and ﬁnd a
2
similar equation for the ycoordinates.
(b) Find the coordinates of P.
19. If f x m0 cm x m has positive radius of convergence and e f x n0 dn x n, show that n P¢ P¶ P£ i ci dn ndn n i 1 i1 FIGURE FOR PROBLEM 18 20. Rightangled 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 fx 1 x
x x2 fn x n is
n1 where fn is the nth Fibonacci number, that is, f1 1, f2 1, and fn fn 1 fn 2
for n 3. [Hint: Write x 1 x x 2
c0 c1 x c2 x 2 . . . and multiply both sides
of this equation by 1 x x 2.]
(b) By writing f x 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 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.
 Fall '09
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