60 Pages

# chapter4

Course Number: MAT 215, Fall 2008

College/University: Princeton

Word Count: 14343

Rating:

###### Document Preview

Chapter 4 Sequences and SeriesFirst View Recall that, for any set A, a sequence of elements of A is a function f : M A. Rather than using the notation f n for the elements that have been selected from A, since the domain is always the natural numbers, we use the notational convention an f n and denote sequences in any of the following forms: an * 1 n an n+M or a1 a2 a3 a4 This is the only time that we...

##### Unformatted Document Excerpt
Coursehero >> New Jersey >> Princeton >> MAT 215

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

4 Chapter Sequences and SeriesFirst View Recall that, for any set A, a sequence of elements of A is a function f : M A. Rather than using the notation f n for the elements that have been selected from A, since the domain is always the natural numbers, we use the notational convention an f n and denote sequences in any of the following forms: an * 1 n an n+M or a1 a2 a3 a4 This is the only time that we use the set bracket notation in a different context. The distinction is made in the way that the indexing is communicated . For an :, the an * 1 is the constant sequence that lists the term : innitely often, n : : : : while an : n + M is the set consisting of one element :. (When you read the last sentence, you should have come up with some version of For a sub n equal to :, the sequence of a sub n for n going from one to innity is the constant sequence that lists the term : innitely often, : : : while the set consisting of a sub n for n in the set of positive integers is the set consisting of one element : i.e., the point is that you should not have skipped over the an * 1 n and an : n + M.) Most of your previous experience with sequences has been with sequences of real numbers, like | }* 3 1 1 2 3 5 8 13 21 34 55 ,... n 1n1 | n2 3n 5 n 47 }* n1 | n3 1 n3 1 }* 1 n n1 | and log n n sin r nH s}* 8 n1 . In this chapter, most of our sequences will be of elements in Euclidean n-space. In MAT127B, our second view will focus on sequence of functions. 123 124 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW As children, our rst exposure to sequences was made in an effort to teach us to look for patterns or to develop an appreciation for patterns that occur naturally. Excursion 4.0.1 For each of the following, nd a description for the general term as a function of n + M that ts the terms that are given. 1. 2 4 8 16 32 64 5 7 9 11 13 15 7 11 3 2. 1 9 81 729 5 9 13 ***An equation that works for (1) is 2n 2n 31 while (2) needs a different formula for the odd terms and the even terms one pair that works is 2n 1 2n 11 for n even and 3n1 when n is odd.*** As part of the bigger picture, pattern recognition is important in areas of mathematics or the mathematical sciences that generate and study models of various phenomena. Of course, the models are of value when they allow for analysis and/or making projections. In this chapter, we seek to build a deeper mathematical understanding of sequences and series primary attention is on properties associated with convergence. After preliminary work with sequences in arbitrary metric spaces, we will restrict our attention to sequences of real and complex numbers. 4.1 Sequences and Subsequences in Metric Spaces If you recall the denition of convergence from your frosh calculus course, you might notice that the denition of a limit of a sequence of points in a metric space merely replaces the role formerly played by absolute value with its generalization, distance. 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES 125 Denition 4.1.1 Let pn * 1 denote a sequence of elements of a metric space S d n and p0 be an element of S. The limit of pn * 1 is p0 as n tends to (goes to or apn proaches) innity if and only if e d 0 " 2M M M + MF 1n n M " d pn p0 1 + U F We write either pn p0 or lim pn n* p0 . Remark 4.1.2 The description M M indicates that limit of sequence proofs require justication or specication of a means of prescribing how to nd an M that will work corresponding to each 0. A function that gives us a nice way to specify M s is dened by z{ 1 and is sometimes referred to as the ceiling function. Note, for example, that 2 1, L22M 2, and L5M 5. Compare this to the greatest integer function, which is sometimes referred to as the oor function. LxM inf j + ] : x n j | }* 2 Example 4.1.3 The sequence has the limit 0 in U. We can take M 1 2, t u t nu n 1 z { 1 3 700 M 200, and M 234. Of course, three examples 100 350 3 z{ 2 . Then n M does not a proof make. In general, for 0, let M implies that n z{ 2 o 2 0 nn n2n n n . nn n 1 2 which, by Proposition 1.2.9 (#7) and (#5), implies that and n 2 n Using the denition to prove that the limit of a sequence is some point in the metric space is an example of where our scratch work towards nding a proof might 126 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW be quite different from the proof that is presented. This is because we are allowed to work backwards with our scratch work, but we can not present a proof that starts with the bound that we want to prove. We illustrate this with the following excursion. Excursion 4.1.4 After reading the presented } scratch work, ll in what is missing to | 1 in * converges to i in F. complete the proof of the claim that n 1 n1 (a) Scratch 1 lim n* n n n1 n nn work towards a proof. Because i + F, it sufces to show that in i. Suppose 0 is given. Then 1 T T nn nn n n n 1 in i n 1 n n 1 i n in 2 2 nn n in n nn n nn 1n n 1 n 1 n1 T T 2 2 whenever n. So taking M will work. 0, let M T implies that n T n and 2 1 (b) A proof. For . Then n + M and n M 2 which is equivalent to T 2 3 2 4 . Because n 1 0, we also know that n . Consequently, if n M , then n nn n1 in n n1 n in n nn 1 nn Since v 1 i.e., 1 i yields that n | 6 i n i n n1 n 1 n n n n n n1i n n n nn 1n T 2 n1 n n1 M "n nn 5 0 was arbitrary, we conclude that t t 0 " 2M M + MF 1n n . Finally, i in 1 }* n1 n uuw n in in n 1 1 n* n in 1 0 1 + F and lim converges to i in F. 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES ***Acceptable responses are (1) 1 n* n lim in 1 i .*** T 2 T 2 , (2) , (3) n n T 1, (4) 2, (5) 127 T 2 , (6) n Denition 4.1.5 The sequence pn * 1 of elements in a metric space S is said to n converge (or be convergent) in S if there is a point p0 + S such that lim pn p0 n* it is said to diverge in S if it does not converge in S . Remark 4.1.6 Notice that a sequence in a metric space S will be divergent in S if | }* 2 its limit is a point that is not in S. In our previous example, we proved that nn1 | }* 2 converges to 0 in U consequently, is convergent in Euclidean 1-space. nn1 | }* b c 2 x + U : x 0 d where d On the other hand, is divergent in U nn1 denotes the Euclidean metric on U, d x y x y. Our rst result concerning convergent sequences is metric spaces assures us of the uniqueness of the limits when they exist. Lemma 4.1.7 Suppose pn * 1 is a sequence of elements in a metric space S d. n Then rK L s p F lim pn q " q p. 1 p 1q p q + S F lim pn n* n* Space for scratch work. Excursion 4.1.8 Fill in what is missing in order to complete the proof of the lemma. Proof. Let pn * 1 be a sequence of elements in a metric space S d for n which there exists p and q in S such that lim pn p and lim pn q. Suppose n* n* 128 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW p and 1 d p q. Because lim pn the p / q. Then d p q 0 and we let n* 2 0, there exists a positive integer M1 such that n similarly, lim pn n* M1 " d pn p q yields the existence of a positive integer M2 such that . 1 Now, let M max M1 M2 . It follows from the symmetry property and the triangular inequality for metrics that n M implies that d p q n d p pn 2 2 3 d p q which contradicts the trichotomy law. Since we have reached a contradiction, we conclude that as needed. Therefore, the limit of any convergent sequence 4 in a metric space is unique. ***Acceptable ll-ins are: (1) n 1 d p q, (4) p q.*** 2 M2 " d pn q , (2) d pn q (3) Denition 4.1.9 The sequence pn * 1 of elements in a metric space S d is n bounded if and only if d e 2M 2x M 0 F x + S F 1n n + M " d x pn M . Note that if the sequence pn * 1 of elements in a metric space S is an not n bounded, then the sequence is divergent in S. On the other hand, our next result shows that convergence yields boundedness. Lemma 4.1.10 If the sequence pn * 1 of elements in a metric space S d is conn vergent in S, then it is bounded. 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES Space for scratch work. 129 Proof. Suppose that pn * 1 is a sequence of elements in a metric space S d n that is convergent to p0 + S. Then, for 1, there exists a positive integer M M 1 such that n M " d pn p0 1. jb c k Because d p j p0 : j + M F 1 n j n M is a nite set of nonnegative real numbers, it has a largest element. Let j jb c kk K max 1 max d p j p0 : j + M F 1 n j n M . Since d pn p0 n K , for each n + M, we conclude that pn * 1 is bounded. n Remark 4.1.11 To see that the converse of Lemma 4.1.10 is false, for n + M , let !1 ! , if 2 n !2 n pn . ! ! ! 1 1 , if 2 0 n n3 Then, for d the Euclidean metric on U1 , d 0 pn pn * 1 is not convergent in U. n Excursion 4.1.12 For each n + M, let an dened in Remark 4.1.11. (a) Use the denition to prove that lim an n* 0 pn 1 for all n + M, but p2n1 where pn is p2n and bn 0. 130 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW 1. (b) Use the denition to prove that lim bn n* if you 3 2 n 1 2n 1 1 used 2 and 1 , respectively, your choices for corresponding M will be n n3 slightly off. The following are acceptable solutions, which of course are not unique compare what you did for general sense and content. Make especially certain that you did not offer a proof that is working backwards from what you wanted to z { 1 show. (a) For 0, let M M M implies that T . Then n 2 n n n1 n b T c1 T 1 1 1 1 n or . If follows that n 0n n 2 n 2 2 2n 2n 2n 2n 2n TT whenever n M. Since 0 was arbitrary, we conclude that z{ 1 1 lim an lim 0. (b) For 0, let M M . Then n M 2 n* n* 2n ***Note that an 2 1 and bn 1 1 2n 1 1 1 1 implies that n 0 implies that 1 or . Note that, for n + M, n o 1 n n 2 0 2 2 0 and 2n 2 n n 2 0 n n. Thus, for n + M and n M, we have that nt nn n u n nn1n 1 1 1 n 1 n 1n n n n n 2n 2 n 2n 2 n . 2n 2 t u 1 Since 0 was arbitrary, we conclude that lim bn lim 1 n* n* 2n 2 1.*** Remark 4.1.13 Hopefully, you spotted that there were some extra steps exhibited in our solutions to Excursion 4.1.12. I chose to show some of the extra steps that 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES 131 illustrated where we make explicit use of the ordered eld properties that were discussed in Chapter 1. In particular, it is unnecessary for you to have explicitly demonstrated that 2n 2 n from the inequalities that were given in Proposition TT 1 1 1.2.9 or the step that was shown in part (a). For the former 2n 2n you can just write things like 2n 2 n for the latter, you could just have written bT c2 1 . 2 2n What we just proved about the sequence given in Remark 4.1.11 can be translated to a statement involving subsequences. Denition 4.1.14 Given a sequence pn * 1 of elements in a metric space X and n k a sequence n kk * 1 of positive integers such that of n k n kj 1 for each k + M, the j k* * sequence pn k k 1 is called a subsequence of pn * 1 . If pn k k 1 converges in n X then its limit is called a subsequential limit of pn * 1 . n Remark 4.1.15 In our function terminology, a subsequence of f : M X is the restriction of f to any innite subset of M with the understanding that ordering is conveyed by the subscripts i.e., n j n j 1 for each j + M. From Excursion 4.1.12, we know that the sequence pn * 1 given in Remark n 4.1.11 has two subsequential limits namely, 0 and 1. The uniqueness of the limit of a convergent sequence leads us to observe that every subsequence of a convergent sequence must also be convergent to the same limit as the original sequence. Consequently, the existence of two distinct subsequential limits for pn * 1 is an n alternative means of claiming that pn * 1 is divergent. In fact, it follows from n the denition of the limit of a sequence that innitely many terms outside of any neighborhood of a point in the metric space from which the sequence is chosen will eliminate that point as a possible limit. A slight variation of this observation is offered in the following characterization of convergence of a sequence in metric space. Lemma 4.1.16 Let pn * 1 be a sequence of elements from a metric space X d. n Then pn * 1 converges to p + X if and only if every neighborhood of p contains n all but nitely many of the terms of pn * 1 . n 132 Space for scratch work. CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Proof. Let pn * 1 be a sequence of elements from a metric space X d. n Suppose that pn * 1 converges to p + X and V is a neighborhood of n p. Then there exists a positive real number r such that V Nr p. From the denition of a limit, there exists a positive integer M M r such that n M implies that d p pn r i.e., for all n M, pn + V . Consequently, at most the pk : k + M F 1 n k n M is excluded from V . Since V was arbitrary, we conclude that every neighborhood of p contains all but nitely many of the terms of pn * 1 . n Suppose that every neighborhood of p contains all but nitely many of the terms of pn * 1 . For any 0, N p contains all but nitely many of the n * terms of pn n 1 . Let M max k + M : pk + N p. Then n M implies that pn + N p from which it follows that d pn p . Since 0 was arbitrary, we conclude that, for every 0 there exists a positive integer M M such that n M implies that d pn p that is, pn * 1 converges to p + X. n It will come as no surprise that limit point of subsets of metric spaces can be related to the concept of a limit of a sequence. The approach used in the proof of the next theorem should look familiar. Theorem 4.1.17 A point p0 is a limit point of a subset A of a metric space X d if and only if there is a sequence pn * 1 with pn + A and pn / p0 for every n n such that pn p0 as n * Proof. ! Suppose that there is a sequence pn * 1 such that pn + A, pn / n p0 for every n, and pn p0 . For r 0, consider the neighborhood Nr p0 . Since pn p0 , there exists a positive integer M such that d pn p0 r for all n M. In particular, p M 1 + A D Nr p0 and p M 1 / p0 . Since r 0 was arbitrary, we conclude that p0 is a limit point of the set A. " Suppose that p0 + X is a limit point of A. (Finish this part by rst making | }* 1 judicious use of the real sequence to generate a useful sequence pn * 1 n j j1 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES followed by using the fact that to p0 .) 133 1 0 as j * to show that pn * 1 converges n j Remark 4.1.18 Since Theorem 4.1.17 is a characterization for limit points, it gives us an alternative denition for such. When called upon to prove things related to limit points, it can be advantageous to think about which description of limit points would be most fruitful i.e., you can use the denition or the characterization interchangeably. We close this section with two results that relate sequences with the metric space properties of being closed or being compact. Theorem 4.1.19 If pn * 1 is a sequence in X and X is a compact subset of a n metric space S d, then there exists a subsequence of pn * 1 that is convergent n in X. Space for scratch work. Proof. Suppose that pn * 1 is a sequence in X and X is a compact subset of n pn : n + M. If P is nite, then there is at least a metric space S d. Let P one k such that pk + P and, for innitely many j + M, we have that p j pk . 134 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW j k* Consequently, we can choose a sequence n j j 1 such that n j n j 1 and pn j k j k* pk for each j + M. It follows that pn j j 1 is a (constant) subsequence of pn * 1 n that is convergent to pk + X. If P is innite, then P is an innite subset of a compact set. By Theorem 3.3.46, it follows that P has a limit point p0 in X. From Theorem 4.1.17, we conclude that there is a sequence qk * 1 with qk + P and k qk / p0 for every k such that qk p0 as k * that is, qk * 1 is a subsequence k of pn * 1 that is convergent to p0 + X. n Theorem 4.1.20 If pn * 1 is a sequence in a metric space S d, then the set of n all subsequential limits of pn * 1 is a closed subset of S. n Space for scratch work. Proof. Let E ` denote the set of all subsequential limits of the sequence pn * 1 n of elements in the metric space S d. If E ` is nite, then it is closed. Thus, we can assume that E ` is innite. Suppose that * is a limit point of E ` . Then, corresponding to r 1, there exists x / * such that x + N1 * D E ` . Since ` , we can nd a subsequence of p * that converges to x. Hence, we x+E n nb 1 c b c can choose n 1 + M such that pn 1 / * and d pn1 * 1. Let = d pn 1 * . Because = 0, * is a limit point of E ` , and E ` is innite, there exists y / * that is in N=4 * D E ` . Again, y + E ` leads to the existence of a subsequence of pn * 1 that converges to y. This allows us to choose n 2 + M such that n 2 n b c b c = n 1 and d pn 2 y . From the triangular inequality, d * pn 2 n d * y 4 b c = d y pn 2 . We can repeat this process. In general, if we have chosen the 2 increasing nite sequence n 1 n 2 n j , then there exists a u such that u / * and = u + Nr j * D E ` where r j . Since u + E ` , u is the limit of a subsequence j1 2 b c of pn * 1 . Thus, we can nd n j 1 such that d pn j1 u r j from which it n follows that b c b c = d * pn j1 n d * u d u pn j1 2r j . 2j j k* The method of selection of the subsequence pn j j 1 ensures that it converges to *. Therefore, * + E ` . Because * was arbitrary, we conclude that E ` contains all of its limit points i.e., E ` is closed. 4.2. CAUCHY SEQUENCES IN METRIC SPACES 135 4.2 Cauchy Sequences in Metric Spaces The following view of proximity of terms in a sequence doesnt isolate a point to serve as a limit. Denition 4.2.1 Let pn * 1 be an innite sequence in a metric space S d. Then n pn * 1 is said to be a Cauchy sequence if and only if n d e 0 " 2M M M + MF 1m 1n n m M " d pn pm . 1 Another useful property of subsets of a metric space is the diameter. In this section, the term leads to a characterization of Cauchy sequences as well as a sufcient condition to ensure that the intersection of a sequence of nested compact sets will consist of exactly one element. Denition 4.2.2 Let E be a subset of a metric space X d. Then the diameter of E, denoted by diam E is sup d p q : p + E F q + E . j Q B k 2 x2 1 and Example 4.2.3 Let A 2 x 1 x2 + U2 : x1 R x1 x2 + U2 : max x1 x2 n 1 2 and diam B T 2 2. Then, in Euclidean 2-space, diam A 136 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW bc Note that, for the sets A and B given in Example 4.2.3, diam A 2 T bc diam A and diam B 2 2 diam B. These illustrate the observation that is made with the next result. Lemma 4.2.4 If E is any subset of a metric space X, then diam E bc diam E . Excursion 4.2.5 Use the space provided to ll in a proof of the lemma. (If you get stuck, a proof can be found on page 53 of our text. The property of being a Cauchy sequence can be characterized nicely in terms of the diameter of particular subsequence. Lemma 4.2.6 If pn * 1 is an innite sequence in a metric space X and E M is the n subsequence p M p M 1 p M 2 , then pn * 1 is a Cauchy sequence if and only n if lim diam E M 0. M* Proof. Corresponding to the innite sequence pn * 1 in a metric space X d n let E M denote the subsequence p M p M 1 p M 2 . Suppose that pn * 1 is a Cauchy sequence. For j + M, there exists a n 1 M ` implies that d pn pm . positive integer M ` M ` such that n m j j j j 1 Let M j M ` 1. Then, for any u ) + E M j , it follows that d u ) . j j j k b c 1 1 Hence, sup d u ) : u + E M j F ) + E M j n i.e., diam E M j n . Now j j 4.2. CAUCHY SEQUENCES IN METRIC SPACES given any 137 1 . For 0, there exists M ) such that j M ) implies that j j k b c M max M j M ) and j M, diam E M j . Since 0 was arbitrary, we conclude that lim diam E M 0. Suppose that lim diam E M M* M* 0 and let 0. Then there exists a positive integer K such that m K implies that diam E m i.e., sup d u ) : u + E m F ) + Em In particular, for n j m we can write n m x and j m y for some positive integers x and y and it follows that c b d pn p j n sup d u ) : u + E m F ) + Em . Thus, we have shown that, for any 0, there exists a positive integer m such that b c n j m implies that d pn p j . Therefore, pn * 1 is a Cauchy sequence. n With Corollary 3.3.44, we saw that any nested sequence of nonempty compact sets has nonempty intersection. The following slight modication results from adding the hypothesis that the diameters of the sets shrink to 0. Theorem 4.2.7 If K n * 1 is a nested sequence of nonempty compact subsets of a n metric space X such that ? n+M n* lim diam K n 0, then K n consists of exactly one point. Space for scratch work. Proof. Suppose that K n * 1 is a nested sequence of nonempty compact subn sets of a metric space X d such that lim diam K n 0. From Corollary 3.3.44, K n * 1 being a nested sequence of nonempty compact subsets implies that n ? K n / 3. n+M n* 138 If ? CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW K n consists of more that one point, then there exists points x and y n+M ? ? in X such that x + Kn , y + K n and x / y. But this yields that n+M n+M 0 d x y n sup d p q : p + K n F q + K n for all n + M i.e., diam E M 0. Because lim diam K n n* d x y for any M + M. Hence, lim diam E M / n* ? 0, it follows immediately that K n consists of n+M exactly one point. Remark 4.2.8 To see that a Cauchy sequence in an arbitrary metric space need not converge to a point that is in the space, consider the metric space S d where S is the set of rational numbers and da b a b. On the other hand, a sequence that is convergent in a metric space is Cauchy there. Theorem 4.2.9 Let pn * 1 be an innite sequence in a metric space S d. If n pn * 1 converges in S, then pn * 1 is Cauchy. n n Proof. Let pn * 1 be an innite sequence in a metric space S d that conn verges in S to p0 . Suppose 0 is given. Then, there exists an M + M such that n M " d pn p0 . From the triangular inequality, if n M and m M, 2 then d pn pm n d pn p0 d p0 pm 2 2 . Since 0 was arbitrary, we conclude that pn * 1 is Cauchy. n As noted by Remark 4.2.8, the converse of Theorem 4.2.9 is not true. However, if we restrict ourselves to sequences of elements from compact subsets of a metric space, we obtain the following partial converse. Before showing this, we will make some us Theorem 4.2.10 Let A be a compact subset of a metric space S d and pn * 1 n be a sequence in A. If pn * 1 is Cauchy, then there exists a point p0 + A such that n pn p0 as n *. 4.3. SEQUENCES IN EUCLIDEAN K -SPACE 139 Proof. Let A be a compact subset of a metric space S d and suppose that j k* pn * 1 of elements in A is Cauchy. Let E M be the subsequence p M j j 0 . n k* j k* j Then E M m 1 is a nested sequence of closed subsets of A and E M D A m 1 is a b c nested sequence of compact subsets of S for which lim diam E M D A 0. By Theorem 4.2.7, there exists a unique p such that p + E M D A for all M. Now justify that pn p as n *. M* 4.3 Sequences in Euclidean k-space When we restrict ourselves to Euclidean space we get several additional results including the equivalence of sequence convergence with being a Cauchy sequence. The rst result is the general version of the one for Euclidean n-space that we discussed in class. Lemma 4.3.1 On Uk d, where d denotes the Euclidean metric, let pn x1n x2n x 3n xkn Then the sequence pn * 1 converges to P p1 p2 p3 pk if and only if n x jn p j for each j, 1 n j n k as sequences in U1 . Proof. The result follows from the fact that, for each m, 1 n m n k, Y Xk T k X; b n c2 ; n 2 W nx jn p j n . xmn pm x jn p j n xmn pm n j1 j1 Suppose that 0 is given. If pn * 1 converges to P p1 p2 p3 pk , then n there exists a positive real number M M such than n M implies that Y Xk X; b c2 d pn P W x jn p j . j1 140 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW M, Hence, for each m, 1 n m n k, and for all n xmn pm n d pn P . Since 0 was arbitrary, we conclude that lim xmn n* pm . Conversely, suppose that x jn p j for each j , 1 n j n k as sequences in U1 . Then, for each j, 1 n j n k, there exists a positive integer M j M j such that n M j implies that xmn pm . Let M k 1n jnk max M j . It follows that, for n M, k rs ;n n nx jn p j n k d pn P n k j1 . Because 0 was arbitrary, we have that lim pn P. n* Once we are restricted to the real eld we can relate sequence behavior with algebraic operations involving terms of given sequences. The following result is one of the ones that allows us to nd limits of given sequences from limits of sequences that we know or have already proved elsewhere. Theorem 4.3.2 Suppose that z n * 1 and ?n * 1 are sequences of complex numn n bers such that lim z n S and lim ?n T . Then n* n* (a) lim z n n* n* n* ?n S T (b) lim cz n (c) lim z n ?n t (d) lim n* cS, for any constant c ST d S , provided that 1n n + M " ? n / T e 0 FT / 0. zn ?n u Excursion 4.3.3 For each of the following, ll in either the proof in the box on the left of scratch work (notes) that support the proof that is given. If you get stuck, proofs can be found on pp 49-50 of our text. Proof. Suppose that z n * 1 and ?n * 1 are sequences of complex numbers n n such that lim z n S and lim ?n T . n* n* 4.3. SEQUENCES IN EUCLIDEAN K -SPACE Space for scratch work. Need look at z n ?n S T Know we can make z m S for m M1 2 & ?n T for n M2 2 Go for M max M1 M2 and use Triangular Ineq. Space for scratch work. Need look at cz n cS c z n S Know we can make z m S for m M1 c for c / 0, mention c 0 as separate case. Since z n S, there exists M1 + M such that n M1 implies that z n S 1. Hence, z n S 1 or z n 1 S for all n M1 . Suppose that 0 is given. If T 0, then ?n 0 as n * implies that there exists M ` + M such that ?n whenever n M ` . For 1 S n max M1 M ` , it follows that t u z n ?n ST z n ?n 1 S . 1 S Thus, lim z n ?n 0. If T / 0, then ?n T as n * yields that there exists M2 + M such that ?n whenever n M2 . 2 1 S From z n S, there exists M3 + M such that n M3 " z n T . Finally, for any n max M1 M2 M3 , 2 T z n ?n ST z n ?n z n T z n T ST n z n ?n T ?n z n S T 1 S . 2 1 S 2 T n* 141 (a) (b) Space for scratch work. (c) 142 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Space for scratch work. tu tu 1 zn zn ?n ?n we can just apply the result from (c). (d) The following result is a useful tool for proving the limits of given sequences in U1 . Lemma 4.3.4 (The Squeeze Principle) Suppose that xn * 1 and yn * 1 are sen n quences of real numbers such that lim xn S and lim yn S. If u n * 1 is a n n* n* sequence of real numbers such that, for some positive integer K xn n u n n yn , for all n then lim u n n* K, S. Excursion 4.3.5 Fill in a proof for The Squeeze Principle. Theorem 4.3.6 (Bolzano-Weierstrass Theorem) In Uk , every bounded sequence contains a convergent subsequence. Proof. Suppose that pn * 1 be a bounded sequence in Uk . Then P n Uk , de f pn : n + M is bounded. Since P is a closed and bounded subset of by the * Heine-Borel Theorem, P is compact. Because pn n 1 is a sequence in P a compact subset of a metric space, by Theorem 4.1.19, there exists a subsequence of pn * 1 that is convergent in P. n 4.3. SEQUENCES IN EUCLIDEAN K -SPACE 143 Theorem 4.3.7 (Uk Completeness Theorem) In Uk , a sequence is convergent if and only if it is a Cauchy sequence. Excursion 4.3.8 Fill in what is missing in order to complete the following proof of the Uk Completeness Theorem. Proof. Since we are in Euclidean k-space, by Theorem 1 , we know that any sequence that is convergent in Uk is a Cauchy sequence. Consequently, we only need to prove the converse. Let pn * 1 be a Cauchy sequence in Uk . Then corresponding to 1, n there exists M M 1 + M such that m n M implies that where d denotes the Euclidean metric. In particular, d pn p M 1 1 for all n 2 B M. Let | b max 1 max d p j d M 1n j nM c 1 } . b c Then, for each j + M, d p j d M 1 n B and we conclude that pn * 1 is a n sequence in Uk . From the Theorem, 3 pn : n + M is a compact subset of Because is a Cauchy sequence in a compact metric space, by Theorem 4.2.10, there exists a p0 + pn : n + M such that pn p0 as n *. Since pn * 1 was arbitrary, we concluded that n . Uk . ***Acceptable responses are: (1) 4.2.9, (2) d pn pm 1, (3) bounded, (4) HeineBorel, and (5) every Cauchy sequence in Uk is convergent.*** From Theorem 4.3.7, we know that for sequences in Uk , being Cauchy is equivalent to being convergent. Since the equivalence can not be claimed over arbitrary metric spaces, the presence of that property receives a special designation. Denition 4.3.9 A metric space X is said to be complete if and only if for every sequence in X, the sequence being Cauchy is equivalent to it being convergent in X. Remark 4.3.10 As noted earlier, Uk is complete. In view of Theorem 4.2.10, any compact metric space is complete. Finally because every closed subset of a metric space contains all of its limit points and the limit of a sequence is a limit point, we also have that every closed subset of a complete metric space is complete. 5 4 * pn n 1 144 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW It is always nice to nd other conditions that ensure convergence of a sequence without actually having the nd its limit. We know that compactness of the metric space allows us to deduce convergence from being Cauchy. On the other hand, we know that, in Uk , compactness is equivalent to being closed and bounded. From the Bolzano-Weierstrass Theorem, boundedness of a sequence gives us a convergent subsequence. The sequence i n * 1 of elements in F quickly illustrates that n boundedness of a sequence is not enough to give us convergence of the whole sequence. The good news is that, in U1 , boundedness coupled with being increasing or decreasing will do the job. Denition 4.3.11 A sequence of real numbers xn * 1 is n (a) monotonically increasing if and only if 1n n + M " x n n xn (b) monotonically decreasing if and only if 1n n + M " x n o xn 1 and 1 . Denition 4.3.12 The class of monotonic sequences consists of all the sequences in U1 that are either monotonically increasing or monotonically decreasing. t u n 1n n 1! o1 Example 4.3.13 For each n + M, . It follows that n n 1 n! n! n 1! o . n n n 1 n 1n is monotonically decreasing. n1 n! Consequently, nn | }* Theorem 4.3.14 Suppose that xn * 1 is monotonic. Then xn * 1 converges if n n and only if xn * 1 is bounded. n Excursion 4.3.15 Fill in what is missing in order to complete the following proof for the case when xn * 1 is monotonically decreasing. n Proof. By Lemma 4.1.10, if xn * 1 converges, then . n Now suppose that is monotonically decreasing and bounded. Let xn : n + M. If P is nite, then there is at least one k such that xk + P and, P xk . On the other hand we have that for innitely many j + M, we have that x j xk m o xk m 1 for all m + M. It follows that xn * 1 is eventually a constant n sequence which is convergent to xk . If P is innite and bounded, then from the xn * 1 n 1 4.3. SEQUENCES IN EUCLIDEAN K -SPACE greatest lower bound property of the reals, we can let g greatest lower bound, 1n n + M " 2 145 inf P. Because g is the . Suppose that g n xM g Because xn * 1 is n 0 is given. Then there exists a positive integer M such that otherwise, . 3 , the transitivity of less than or equal to yields 4 that, for all n M, g n x n g was arbitrary, we conclude that . Hence, n 5 M " xn g . Since . 0 ***Acceptable responses are: (1) it is bounded, (2) g n xn , (3) g lower bound that is greater than g, (4) decreasing, and (5) lim xn n* would be a g.*** 4.3.1 Upper and Lower Bounds Our next denition expands the limit notation to describe sequences that are tending to innity or negative innity. Denition 4.3.16 Let xn * 1 be a sequence of real numbers. Then n (a) xn * as n * if and only if r 1K K + U1 " 2M M + MF 1n n and (b) xn * as n * if and only if r 1K K + U1 " 2M M + MF 1n n In the rst case, we write lim x n n* lim xn *. s M " xn n K s M " xn o K * and in the second case we write n* 146 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Denition 4.3.17 For xn * 1 be a sequence of real numbers, let E denote the set n of all subsequential limits in the extended real number system (this means that * and/or * are included if needed). Then the limit superior of xn * 1 is x ` n sup E and the limit inferior of xn * 1 is x` inf E. n We will use lim supxn to denote the limit superior and lim infxn to denote the limit inferior of n* xn * 1 . n n* Example 4.3.18 For each n + M, let an and lim infan n* 1 1n 1 . Then the lim supan 2n n* 2 0. Excursion 4.3.19 Find the limit superior and the limit inferior for each of the following sequences. | 1. sn n 1n 2n n 1 }* n1 2. Q sn 1n 1 sin H n R* 4 n1 | 3. sn t 1 u 1r 1 n Hn s sin 2 }* n1 4.3. SEQUENCES IN EUCLIDEAN K -SPACE 4. Q sn n 4 On P 4 1n R* n1 147 ***For (1), we have two convergent subsequences to consider s2n 3 while s2n1 1 and you should have concluded that lim supsn 3 and lim infsn 1. In working on (2), you should have gotten 5 subsequential limits: s4k 1, T T 2 2 s4k 1 and s4k 3 give two subsequential limits, 1 for k even and 1 for 2 2 k odd s4k 2 also gives two subsequential limits, 2 for k odd and 0 for k even. Comparison of the 5 subsequential limits leads to the conclusion that lim supsn n* T 2 2. The sequence given in (3) leads to three subsequen1 and lim infsn n* 2 tial limits, namely, 0 1, and 2 which leads to the conclusion that lim supsn 2 and lim infsn 0. Finally, for (4), the subsequences s4k s4k 1 s4k n* 13 3 s4k 3 give limits of 1 , and , respectively hence, lim supsn 42 4 n* 3 lim infsn .*** n* 4 n* 2 , n* n* and 3 and 2 Theorem 4.3.20 Let sn * 1 be a sequence of real numbers and E be the set of n (nite) subsequential limits of the sequence plus possibly * and *. Then (a) lim sup sn + E, and n* tt (b) 1x x lim sup sn n* n* u " 2M n u M " sn x . Moreover, lim sup sn is the only real number that has these two properties. Excursion 4.3.21 Fill in what is missing in order to complete the following proof of the theorem. 148 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Proof. For the sequence of real numbers sn * 1 , let E denote the set of subn sequential limits of the sequence, adjoining * and/or * if needed, and s ` lim sup sn . Proof of part (a): If s ` *, then E is unbounded. Thus sn * 1 is not bounded n j k* above and we conclude that there is a subsequence sn k k 1 of sn * 1 such that n lim sn k *. k* n* *, then sn * 1 has no nite subsequential limits i.e., * is If s ` n the only element of E. It follows that lim sn *. n* Suppose that s ` + U. Then E is bounded above and contains at least one element. By CN Theorem 4.1.20, the set E is . It follows from CN Theorem 2 that s` 1 sup E + E E. Proof of part (b): Suppose that there exists x + U such that x s ` and sn o x for innitely many natural numbers n. Then there exists a subsequence of sn * 1 n that converges to some real number y such that . From the triangular inequality, y s ` . But y + E and y s ` contradicts the denition of ` there are at most nitely many n + M . It follows that, for any x s 4 3 for which 5 . Hence, for any x M implies that sn x. s ` there exists a positive integer M such that n Proof of uniqueness. Suppose that p and q are distinct real numbers that satisfy property (b). Then 1x x and 1x x q " 2K n K " sn x . p " 2M n M " sn x Without loss of generality we can assume that p q. Then there exists * + U such that p * q. Since * p there exists M + M such that n M implies that sn *. In particular, at most nitely many of the sk satisfy . Therefore, q cannot be the limit of any subsequence of sn * 1 from which it follows n that q + E i.e., q does not satisfy property (a). 6 4.4. SOME SPECIAL SEQUENCES 149 ***Acceptable responses are: (1) closed, (2) 3.3.26, (3) y o x, (4) sup E, (5) sn o x, and (6) q sk *.*** Remark 4.3.22 Note that, if sn * 1 is a convergent sequence of real numbers, say n lim sn s0 then the set of subsequential limits is just s0 and it follows that lim sup sn n* lim inf sn . n* Theorem 4.3.23 If sn * 1 and tn * 1 are sequences of real numbers and there n n exists a positive integer M such that n M implies that sn n tn , then lim inf sn n lim inf tn n* n* and lim sup sn n lim sup tn . n* n* Excursion 4.3.24 Offer a well presented justication for Theorem 4.3.23. 4.4 Some Special Sequences This section offers some limits for sequences with which you should become familiar. Space is provided so that you can ll in the proofs. If you get stuck, proofs can be found on page 58 of our text. Lemma 4.4.1 For any xed positive real number, lim 1 n* n p 0. 150 Proof. For CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW 0, let M M t u1 p 1 . Lemma 4.4.2 For any xed complex number such x that x 1, lim x n n* 0. Proof. If x 0, then x n 0 for each n + M and lim x n 0. Suppose that x is n* a xed complex number such that 0 x 1. For 0, let , for o 1 !1 ! { z . M M ! ln ! , for 1 ln x The following theorem makes use of the Squeeze Principle and the Binomial Theorem. The special case of the latter that we will use is that, for n + M and 4.4. SOME SPECIAL SEQUENCES ? + U 1, 1 ? n n ; tn u k0 151 tu n ? , where k k k n! . n k!k! n? and 1 ? n bn c k In particular, if ? 0 we have that 1 each k, 1 n k n n. Theorem 4.4.3 (a) If p n* ? n o 1 T n p 1. ? k for 0, then lim 1. n* (b) We have that lim (c) If p T n n 0 and : + U, then lim n: n* 1 pn 0. Proof of (a). We need prove the statement only for the case of p 1 the result 1 for 0 p 1 will follow by substituting in the proof of the other case. If p 1, p T n p 1. Then x 0 and from the Binomial Theorem, then set xn n 1 and 0 xn n p1 . n nxn n 1 xn n p Proof of (b). Let xn T n n 1. Then x n o 0 and, from the Binomial Theorem, n 1 xn n o n n 1 2 xn . 2 152 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Proof of (c). Let k be a positive integer such that k :. For n tu nk n n 1 n 1 n k 1 k n p p 1 p k k! and 0 n: 1 pn . 2k, n k pk 2k k! 4.5 Series of Complex Numbers For our discussion of series, we will make a slight shift is subscripting namely, it will turn out to be more convenient for us to have our initial subscript be 0 instead of 1. Given any sequence of complex numbers ak * 0 , we can associate (or derive) k 3n a related sequence Sn * 0 where Sn ak called the sequence of nt h partial k0 n sums. The associated sequence allows us to give precise mathematical meaning to the idea of nding an innite sum. 3 Denition 4.5.1 Given a sequence of complex numbers ak * 0 , the symbol * 0 ak k k is called an innite series or simply a series. The symbol is intended to suggest an innite summation 3n and each an is called a term in the series. For each n + MC 0, let Sn k 0 ak a0 a1 an . Then Sn * 0 is called the sequence of nth partial sums for n 3* ak . k0 a0 a1 a2 a3 4.5. SERIES OF COMPLEX NUMBERS 153 On the surface, the idea of adding an innite number of numbers has no real meaning which is why the series has been dened just as a symbol. We use the associated sequence of nth partial sums to create an interpretation for the symbol that is tied to a mathematical operation that is well dened. 3 Denition 4.5.2 An innite series * 0 ak is said to be convergent to the complex k number S if and only if the sequence of nth partial sums Sn * 0 is convergent to n 3 S when this occurs, we write * 0 ak S. If Sn * 0 does not converge, we say k n that the series is divergent. Remark 4.5.3 The way that convergence of series is dened, makes it clear that we really arent being given a brand new concept. In fact, given any sequence Sn * 0 , n 3n there exists a sequence ak * 0 such that Sn ak for every k + MC 0: To k1 k see this, simply choose a0 S0 and ak Sk Sk1 for k o 1. We will treat sequences and series as separate ideas because it is convenient and useful to do so. The remark leads us immediately to the observation that for a series to converge it is necessary that the terms go to zero. Lemma 4.5.4 (kth term test) If the series Proof. Suppose that 3* k 0 ak 3* k 0 ak converges, then lim ak k* 0. S. S. Then lim Sk k* S and lim Sk1 k* Hence, by Theorem 4.3.2(a), k* lim ak k* lim Sk Sk1 k* lim Sk lim Sk1 k* SS 0. Remark 4.5.5 To see that the converse is not true, note that the harmonic series * ;1 k1 k is divergent which is a consequence of the following excursion. 154 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Excursion 4.5.6 Use the Principle of Mathematical Induction to prove that, for 3* 1 n 1 . k 1 k , S2n 2 Excursion 4.5.7 Use the denition of convergence (divergence) to discuss the following series. (a) 3* Hk k 1 sin 4 (b) 3* 1 k 1 kk 1 ***The rst example can be claimed as divergent by inspection, because the nth term does not go to zero. The key to proving that the second one converges is 1 1 1 noticing that in fact, the given problem is an example of kk 1 k k1 what is known as a telescoping sum.*** 4.5. SERIES OF COMPLEX NUMBERS 155 The following set of lemmas are just reformulations of results that we proved for sequences. Lemma 4.5.8 (Cauchy Criteria for Series Convergence) The series (of complex 3 0 there exists a positive numbers) * 0 ak is convergent if and only if for every k integer M M such that 1m 1n m n M " Sm Sn >. Proof. The lemma holds because the complex sequence of nth partial sums Sn * 0 is convergent if and only if it is Cauchy. This equivalence follows from the n combination of Theorem 4.2.9 and Theorem 4.3.6(b). Remark 4.5.9 We will frequently make use of the following alternative formulation for the sequence of nt h partial sums being Cauchy. Namely, Sn * 0 is Cauchy if n and only if for every > 0, there exists a positive integer M such that n M n3 n n np n implies that Sn p Sn n k n 1 ak n >, for p 1 2 . 3 ak Lemma 4.5.10 For3 series (of complex numbers) * 0 ak , let Re 3 xk and the k 3 Im ak yk . Then * 0 ak is convergent if and only if * 0 xk and * 0 yk are k k k convergent (real) sequences. Proof. For the complex series Sn n ; k0 3* k 0 ak , n ; k0 ak n ; k0 yk xk i n ; k0 xk n ; k0 yk . 2. Consequently, the result is simply a statement of Lemma 4.3.1 for the case n 3 Lemma 4.5.11 Suppose that * 0 ak is a series of nonnegative real numbers. k 3* Then k 0 ak is convergent if and only if its sequence of nth partial sums is bounded. 3 Proof. Suppose that * 0 ak is a series of nonnegative real numbers. Then k Sn * 0 is a monotonically increasing sequence. Consequently, the result follows n from Theorem 4.3.14. 3 3 Lemma 4.5.12 Suppose that * 0 u k and * 0 ) k are convergent to U and V , k k respectively., and c is a nonzero constant. Then 156 1. 2. 3* 3* k 0 u k k 0 cu k CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW )k cU . U V and Most of our preliminary discussion of series will be with series for which the terms are positive real numbers. When not all of the terms are positive reals, we rst check for absolute convergence. 3 Denition 4.5.13 The series * 0 a j is said to be absolutely convergent if and j 3 3 3 only if * 0 a j converges. If * 0 a j converges and * 0 a j diverges, then j j j 3 the series * 0 a j is said to be conditionally convergent. j After the discussion of some tests for absolute convergence, we will see that * 3 1n absolute convergence implies convergence. Also, we will justify that is n n1 conditionally convergent. 4.5.1 Some (Absolute) Convergence Tests While the denition may be fun to use, we would like other means to determine convergence or divergence of a given series. This leads us to a list of tests, only a few of which are discussed in this section. Theorem 4.5.14 (Comparison Test) Suppose that numbers). 3* k 0 ak is a series (of complex (a) If there exists a positive integer M such that 1k k o M " ak n ck for 3 3 real constants ck and * 0 ck converges, then * 0 ak converges absolutely. k k (b) If there exists a positive integer M such that 1k3 o M " ak o dk o 0 k 3* for real constants dk and k 0 dk diverges, then * 0 ak diverges. k 3 Proof of (a). Suppose that * 0 ak is a series (of complex numbers), there k 3 exists a positive integer M such that 1k k o M " ak n ck , and * 0 ck conk verges. For xed 0, there exists a positive integer K such that n K and p + M implies that n n np nn p n ;n ; n ck n c >. n nk n 1 n k n 1 k 4.5. SERIES OF COMPLEX NUMBERS For n that M` 157 max M K and any p + M, it follows from the triangular inequality n n np np nn p n ; ; ; n n ak n ak n n ck >. n nk n 1 n k n 1 kn1 3 Since 0 was arbitrary, we conclude that * 0 ak converges. k 3 Proof of (b). Suppose that * 0 ak is a series (of real numbers), there exk 3 ists a positive integer M such j 1k k o M " ak o dk o 0, and * 0 dk that k k* 3n diverges. From Lemma 4.5.11, dk n 0 is an unbounded sequence. Since k0 n ; kM ak o n ; kM dk k* j3n ak n 0 is an unbounded. Therefore, for each n M, it follows that k0 3* ak diverges. k0 In order for the Comparison Tests to be useful, we need some series about which convergence or divergence behavior is known. The best known (or most famous) series is the Geometric Series. 3* Denition 4.5.15 For a nonzero constant a, the series metric series. The number r is the common ratio. k 0 ar k is called a geo- Theorem 4.5.16 (Convergence Properties of the Geometric Series) For a / 0, 3 a the geometric series * 0 ar k converges to the sum whenever 0 r 1 k 1r and diverges whenever r o 1. Proof. The claim will follow upon showing that, for each n + MC 0, n ; k0 ar k b a 1 rn 1r 1 c . 158 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW The proof of the next result makes use of the regrouping process that was applied to our study of the harmonic series. j k* Theorem 4.5.17 If a j j 0 is a monotonically decreasing sequence of nonnegative 3 3 real numbers, then the series * 0 a j is convergent if and only if * 0 2 j a2 j j j converges. Excursion 4.5.18 Fill in the two blanks in order to complete the following proof of Theorem 4.5.17. j k* Proof. Suppose that a j j 0 is a monotonically decreasing sequence of nonnegative real numbers. For each n k + MC 0, let Sn n ; j0 aj and Tk k ; j0 2 j a2 j . j k* Note that, because a j j 0 is a monotonically decreasing sequence, for any j + MC 0 and m + M, m 1 a j o a j aj 1 aj m o m 1 a j m, j k* while a j j 0 a sequence of nonnegative real numbers yields that Sn and Tk are monotonically decreasing sequences. For n 2k , b c Sn n a0 a1 a2 a3 a4 a5 a6 a7 a2k a2k1 1 _ ^] ` _ ^] ` _ ^] ` 21 ter ms 22 terms 2k ter ms 4.5. SERIES OF COMPLEX NUMBERS from which it follows that 159 1 For n Sn o a0 2k , a1 a2 a3 a4 _ ^] ` 21 terms a5 _ a6 22 ^] a7 a8 ` b a k1 _2 1 ^] ter ms a2k c ` ter ms 2k1 from which it follows that 2 The result now follows because we have that Sn and Tk are simultaneously bounded or unbounded. ***For (1), the grouping indicated leads to Sn n a1 a0 2a2 4a4 2k a2k a1 Tk , while the second regrouping yields that Sn o a0 a1 a2 2a4 4a8 1 1 2k1 a2k a0 a1 Tk .*** 2 2 As an immediate application of this theorem, we obtain a family of real series for which convergence and divergence can be claimed by inspection. Theorem 4.5.19 (Convergence Properties of p-series) The series verges whenever p 1 and diverges whenever p n 1. 3* 1 connp n1 0, then | Proof. }If p n 0, the pseries diverges by the kth term test. If p 1* an is a monotonically decreasing sequence of nonnegative real numnp n 1 bers. Note that * ; j0 2 j a2 j * ; j 1 2 j b cp 2j 0 * ;r j0 21 p sj . 160 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW Now use your knowledge of the geometric series to nish the discussion. A similar argument yields the following result with is offered without proof. It is discussed on page 63 of our text. Lemma 4.5.20 The series whenever p n 1. 3* 1 converges whenever p j ln j p 1 and diverges j2 Excursion 4.5.21 Discuss the convergence (or divergence) of each of the following. (a) * ; n1 n n2 1 (b) * ;1 n3 n1 * ; n1 (c) 2n 1 n1 4.5. SERIES OF COMPLEX NUMBERS (d) * ; n1 161 n2 3 3n 1 ***Notice that all of the series given in this excursion are over the positive reals thus, checking for absolute convergence is the same as checking for convergence. At this point, we only the n th term test, Comparison, recognition as a p-series, or rearrangement in order to identify the given as a geometric series. For (a), noticing n n 1 that, for each n + M, 2 o2 allows us to claim divergence n 1 n n n1 by comparison with the shifted harmonic series. The series given in (b) is con1 n1 vergent as a p-series for p 3. Because lim / 0 the series given 2n 1 2 in (c) diverges by the n th term test. Finally, since 3n 1 0 for each n + M, 3 3 n 2 which allows us to claim convergence of the series given in n 2 3n 1 n * ;3 (d) by comparison with which is convergent as a constant multiple times the n2 n1 p-series with p 2.*** When trying to make use of the Comparison Test, it is a frequent occurrence that we know the nature of the series with which to make to comparison almost by inspection though the exact form of a benecial comparison series requires some creative algebraic manipulation. In the last excursion, part (a) was a mild example of this phenomenon. A quick comparison of the degrees of the rational functions that form the term suggest divergence by association with the harmonic sen 1 ries, but when we see that 2 we have to nd some way to manipulate n 1 n n the expression 2 more creatively. I chose to illustrate the throwing more in n 1 the denominator argument as an alternative, note that for any natural number n, n 1 n 2 o 1 " 2n 2 o n 2 1 " 2 o which would have justied divergence n 1 2n by comparison with a constant multiple of the harmonic series. We have a nice variation of the comparison test that can enable us to bypass the need for the algebraic manipulations. We state here and leave its proof as exercise. Theorem 4.5.22 (Limit Comparison Test) Suppose that an * 0 and bn * 0 are n n 162 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW an n* bn L 0. Then either such that an o 0, bn o 0 for each n + MC 0, and lim * * ; ; an and bn both converge or both diverge. n1 n1 We have two more important and well known tests to consider at this point. 3 Theorem 4.5.23 (Ratio Test) The series * 0 ak k n n n ak 1 n n (i) converges absolutely if lim sup n n a n 1 k k* (ii) diverges if there exists a nonnegative integer M such that k n n n ak 1 n n n n a n o 1. k 3* M implies that n n n ak 1 n n n 1. It Proof. Suppose that the series k 0 ak is such that lim sup n ak n k* follows that we can nd a positive real number ; such that ; 1 and there exists n n n ak 1 n n an M + M such that n M implies that n n a n ;. It can be shown by induction k n n that, for each p + M and n M, nan p n ; p an . In particular, for n o M 1 and * ; n n a M 1 ; p is convergent as p + M C 0, nan p n ; p a M 1 . Now, the series a geometric series with ratio less than one. Hence, p1 * ; aj * ; aM p is abso- jM1 lutely convergent by comparison from which it follows that convergent. Suppose that the series integer M for which k 3* k k 0 an 3*p 1 k 0 ak is absolutely is such that there exists a nonnegative n n ak 1 n n M implies that n n a n o 1. Briey justify that this yields k divergence as a consequence of the n th term test. 4.5. SERIES OF COMPLEX NUMBERS 163 n n n ak 1 n n 1 leads to no conclusive information conRemark 4.5.24 Note that lim n k* n ak n 3 cerning the convergence or divergence of * 0 ak . k Example 4.5.25 Use the Ratio Test to discuss the convergence of each of the following: 1. * ; n n n n n n ak 1 n n1 n 1 n n n 1!n n For an 0 as n *. Hence, n n! n na n n kn n n n n ak 1 n n ak 1 n n n lim n lim sup n n ak n k* n ak n 1 and we conclude that the series is (absok* lutely) convergent from the ratio test. 1 n1! , 1 n 1! 1 2. * ; n2 n1 n n n n t u n n 12 2n n n ak 1 n 1 12 n2 1 n n n Let an . Then n 1 as n n 1 2n na n n2 2n nn 2 n 2 k n n n n n ak 1 n n ak 1 n 1 n n n *. Thus, lim sup n lim n 1 and we conclude n ak n n ak n k* 2 k* that the given series is (absolutely) convergent. 3 T Theorem 4.5.26 (Root Test) For * 0 ak , let : lim sup k ak , k k* 2n 3 (i) if 0 n : 1, then * 0 ak converges absolutely k 3* (ii) if : 1, then k 0 ak diverges and (iii) if : 3* 1, then no information concerning the convergence or divergence of ak can be claimed. k0 3 T Proof. For * 0 ak , let : lim sup k ak . If : 1, there exists a real number k k* ; such that : ; 1. Because : is a supremum of subsequential limits and : ; 1, by Theorem 4.3.20, there exists a positive integer M such that n M * ; T implies that n an ; i.e., an ; n for all n M. Since ; j is convergent jM1 164 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW 3 ;m 1 as a geometric series (that sums to ), we conclude that * 0 ak converges k 1; 3* that is, k 0 ak converges absolutely. 3 Briey justify that : 1 leads to divergence of * 0 ak as a consequence k of the n th term test. Finally, since : T lim sup k ak k* 1 for the p-series, we see that no conclusion can be drawn concerning the convergence of divergence of the given series. * ;n Example 4.5.27 Use the Root Test, to establish the convergence of . 2n1 n1 T From Theorem 4.4.3(a) and (b), lim n 2n 1. Hence, U lim sup k k 2k1 k* Vt u k lim k 2 k k* 2 n* T k k* lim 2k 2 1 1 2 from which we claim (absolute) convergence of the given series. Thus far our examples of applications of the Ration and Root test have led us n n n n ak 1 n n ak 1 n k n n or lim supTak to exam sequences for which lim sup n lim n na n na n k* k k k* k* T lim k ak . This relates back to the form of the tests that you should have seen with k* your rst exposure to series tests, probably in frosh (or AP) calculus. Of course, the point of offering the more general statements of the tests is to allow us to study the absolute convergence of series for which appeal to the limit superior is necessary. The next two excursion are in the vein the parts that are described seek to help you to develop more comfort with the objects that are examined in order to make use of the Ratio and Root tests. 4.5. SERIES OF COMPLEX NUMBERS t u !1ij ! ! ! ! 2 ! t uj !2 ! ! ! 5 , if 2 j . , if 2 0 j 165 Excursion 4.5.28 For n + MC 0, let a j n} |n na j 1 n * n n 1. Find the rst four terms of n . aj n j 0 2. Find the rst four terms of QTn nR* jn ajn . j1 n} |n naj 1 n * n n 3. Find E 1 the set of subsequential limits of n aj n j 0 QTn nR* 4. Find E 2 the set of subsequential limits of j na j n j1 5. Find each of the following: n} |n na j 1 n * n (a) lim sup n na n j j* j0 166 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW n} |n na j 1 n * n n (b) lim inf n j * aj n j 0 (c) lim sup j* QTn nR* jn ajn j1 (d) lim inf j * Q Tn nR* jn ajn j1 * ; 6. Discuss the convergence of aj j0 } 2 5 16 125 ***For (1), we are looking at while (2) is 5 4 125 32 T T n n na j 1 n 2 22 2 n n for (3), if c j n a j n, then the possible subsequential 5252 jk j k limits are given by looking at c2 j and c2 j1 and E 1 0 * if in (4) we let T Tn n jk j k 22 jn a j n, then consideration of d2 j and d2 j1 leads to E2 dj 25 T 2 2 For (3) and (4), we conclude that the requested values are *, 0, , and , re2 5 spectively. For the discussion of (6), note that The Ratio Test yields no information because neither (a) nor (b) is satised in the other hand, from (5c), we see that T Q Tn nR* 2 jn lim sup ajn 1, from which we conclude that the given series j1 2 j* is absolutely convergent. (As an aside, examination of S2n and S2n1 corre- | 4.5. SERIES OF COMPLEX NUMBERS sponding to 4 1 2i 10 21 167 * ; a j even allows us to conclude that the sum of the given series is j0 134 168i .)*** 105 t u ! 2j ! ! ! !3 1 , if 2 j . , if 2 0 j Excursion 4.5.29 For n + MC 0, let a j ! t u j1 !2 ! ! ! 3 n} |n na j 1 n * n . 1. Find the rst four terms of n na n j j0 2. Find the rst four terms of QTn nR* jn ajn . j1 n} |n naj 1 n * n n 3. Find E 1 the set of subsequential limits of n aj n j 0 QTn nR* 4. Find E 2 the set of subsequential limits of j na j n j1 5. Find each of the following: 168 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW n} |n na j 1 n * n n (a) lim sup n aj n j 0 j* n} |n na j 1 n * n n (b) lim inf n j * aj n j 0 (c) lim sup j* QTn nR* jn ajn j1 (d) lim inf j * Q Tn nR* jn ajn j1 * ; 6. Discuss the convergence of aj j1 tu tu 3 233 23 ***Response this time are: (1) , 23 23 tu U tu U t u54 U | } 2 32 2 2 2 23 42 442 34 3 (3) E 1 , (2) 1 3 33 3 93 93 29 |} Q Tn nR 2 2j n (4) E2 where this comes from separate consideration of lim a2 j n j* 3 Q Tn nR 342 2 2 j1 n a2 j1 n , (5) the values are and , respectively. Finally, and lim j* 293 3 the Ratio Test fails to offer information concerning convergence, however, the Root 4.5. SERIES OF COMPLEX NUMBERS Test yields that 169 * ; a j is absolutely convergent. (Again, if we choose to go back to j1 the denition, examination of the nt h partial sums allows us to conclude that the series converges to 3.)*** n n n ak 1 n 3 n n 1 for a series * 0 ak , the ratio Remark 4.5.30 Note that, if lim sup n k n ak k* test yields no information concerning the convergence of the series. 4.5.2 Absolute Convergence and Cauchy Products When the terms in the generating sequence for a series are not all nonnegative reals, we pursue the possibility of different forms of convergence. The next result tells us that absolute convergence is a stronger condition than convergence j k* 3 Lemma 4.5.31 If a j j 1 is a sequence of complex numbers and * 1 a j conj n3 n3 3* n* n * verges, then j 0 a j converges and n j 0 a j n n j 0 a j . Proof. (if we were to restrict ourselves to real series) The following argument that is a very slight variation of the one offered by the author of our text applies only to series over the reals it j followed by a general argument that applies to series of is k 3 * complex terms. Suppose a j j 1 is a sequence of real numbers such that * 1 a j j converges and dene )j Then ) j * j a j 2 *j aj and * j a j a j 2 a j . Furthermore, aj a j and * j 0 a j while ) j a j o 0 implies that ) j while a j 0 implies that ) j 0 and * j a j a j Consequently, 0 n ) n n an and 0 n u n n an and, from the Comparison Test, it b c 3 3 3 follows that * 0 ) j and * 0 * j converge. By Lemma 4.5.12, * 0 ) j * j j j j converges. Finally, since b cb cb c ) j *j n ) j *j n ) j *j , 170 we see that * ;b j1 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW )j *j n c * ;b j1 ) j *j n c * ;b j1 )j c *j n n nn 3* n n 3* 3* n n na j n n na j n. Hence 0 n n3* a j n n 3* na j n i.e., j 1 aj n j 1 n j1 n j1 j1 The following proof of the lemma in general makes use of the Cauchy Criteria for Convergence. j k* Proof. Suppose that a j j 1 is a sequence of complex numbers such that 3* 0 is given. Then there exists a positive integer M j 1 a j converges and 3m M such that 1m 1n m n M " Sm Sn > where Sm a . n3 n j1 j 3n p n np n In particular, for any p + M and n M, j n 1 a j n j n 1 a j n . From n3 n3 n np n np the triangular inequality, it follows that n j n 1 a j n n for any j n 1 a j 3* p + M and n M. Since 0 was arbitrary, we conclude that j 1 a j converges by the Cauchy Criteria for Convergence. Remark 4.5.32 A re-read of the comparison, root and ratio tests reveals that they are actually tests for absolute convergence. Absolute convergence offers the advantage of allowing us to treat the absolutely convergence series much as we do nite sums. We have already discussed the term by term sums and multiplying by a constant. There are two kinds of product that come to mind: The rst is the one that generalizes what we do with the distributive law (multiplying term-by-term and collecting terms), the second just multiplies the terms with the matching subscripts. Denition 4.5.33 (The Cauchy Product) For Ck Then 3* k 0 Ck k ; j0 3* j 0 aj and 3* j 0 bj, set a j bk j for each k + MC 0 . is called the Cauchy product of the given series. 3 3 Denition 4.5.34 (The Hadamard Product) For * 0 a j and * 0 b j , the sej j 3 ries * 0 a j b j is called the Hadamard product of the given series. j 4.5. SERIES OF COMPLEX NUMBERS 171 The convergence of two given series does not automatically lead to the convergence of the Cauchy product. The example given in our text (pp 73-74) takes aj bj 1 j T . j1 3 We will see in the next section that * 0 a j converges (conditionally). On the j 3k 3 1 other hand, Ck is such that 1k k 0 T j 0 a j bk j j k j 1 j 1 C k o k ; j0 2 k 2 k 1 2 k 2 which does not go to zero as k goes to innity. If one of the given series is absolutely convergent and the other is convergent we have better news to report. 3 3 3 Theorem 4.5.35 (Mertens Theorem) For * 0 a j and * 0 b j , if (i) * 0 a j j3 j j 3 converges absolutely, (ii) * 0 a j A, and * 0 b j B, then the Cauchy j 3 3j product of * 0 a j and * 0 b j is convergent to AB. j j 3 3 Proof. For * 0 a j and * 0 b j , let An and Bn be the respective sej j quences of nth partial sums. Then n k ;; Cn a j bn j a0 b0 a0 b1 a1 b0 a0 bn a1 bn1 an b0 k0 j0 172 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW which can be rearrangedusing commutativity, associativity and the distributive lawsto a0 b0 Thus, 3* Cn a0 Bn de f b1 bn a1 b0 b1 bn1 an b0 . a1 Bn1 an1 B1 an B0 . 0. Substitution in an B ;0 Since j 0 bj B, for ;n ;n B Bn we have that lim ;n n* the previous equation yields that Cn a0 B a1 B a0 ;n a0 ;n ;n1 an1 B an1 ;1 ;1 which simplies to Cn Let <n Because lim An n* An B a1 ;n1 an ;0 . a1 ;n1 an1 ;1 an ;0 A, we will be done if we can show that lim <n 0. In view n 3* 3* n n* of the absolute convergence of j 0 a j , we can set j 0 na j n :. Suppose that 0 is given. From the convergence of ;n , there exists a positive integer M such that n M implies that ;n . For n M, it follows that <n anM1 ; M 1 anM ; M nn 3 From the convergence of ;n and * 0 na j n, we have that j a0 ;n anM ; M a1 ;n1 anM1 ; M 1 a0 ;n a1 ;n1 an1 ;1 an ;0 : while M being a xed number and ak 0 as k * yields that an1 ;1 an ;0 0 as n *. Hence, <n a0 ;n a1 ;n1 anM1 ; M 1 anM ; M an1 ;1 an ;0 implies that lim sup <n n :. Since 0 was arbitrary, it follows that lim <n 0 as n* n* needed. The last theorem in this section asserts that if the Cauchy product of two given convergent series is known to converge and its limit must be the product of the limits of the given series. 4.5. SERIES OF COMPLEX NUMBERS 173 3 3 3 Theorem 4.5.36 If the series * 0 a j , * 0 b j , and * 0 c j are known to conj j j 3* 3* 3 verge, aj A, bj B, and * 0 c j is the Cauchy product of j 03 j3 0 j 3* a j and * 0 b j , then * 0 c j AB. j0 j j 4.5.3 Hadamard Products and Series with Positive and Negative Terms 3* 1 j1 can be realized as several different Hadamard prodj j 33 1 1 1 1 ucts letting a j , bj , cj and d j , gives us j j j 3 j 33 j 32 3* 3 3 1 as the Hadamard product of * 1 a j and * 1 b j as well as the j1 j j 3 j j 3 3* 3* 3* Hadamard product of j 1 c j and j 1 d j . Note that only j 1 a j diverges. The following theorem offers a useful tool for studying the nt h partial sums for Hadamard products. j k* Theorem 4.5.37 (Summation-by-Parts) Corresponding to the sequences a j j 0 , let An n ; j0 Notice that a j for n + MC 0 , and A1 0. j k* Then for the sequence b j j 0 and nonnegative integers p and q such that 0 n p n q, q ; jp ajbj q1 ; jp b Aj bj bj c 1 Aq bq A p1 b p Excursion 4.5.38 Fill in a proof for the claim. 174 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW As an immediate application of this formula, we can show that the Hadamard product of a series whose nt h partial sums form a bounded sequence with a series that is generated from a monotonically decreasing sequence of nonnegative terms is convergent. Theorem 4.5.39 Suppose that the series (i) Q3 n j 0 aj 3* j 0 aj and 3* j 0 bj are such that R* n0 is a bounded sequence, j k* (ii) b j j 0 is a monotonically decreasing sequence of nonnegative reals, and (iii) lim b j j* 0. is convergent. Then 3* j 0 ajbj 3n integer Proof. For each n + M, let An j 0 a j . Then there exists a positive k j* M such that An n M for all n. Suppose that 0 is given. Because b j j 0 is monotonically decreasing to zero, there exists a positive integer K for which b K . Using summation-by-parts, for any integers p and q satisfying K n q n p, it 2K follows that n3 n n n b c nq n n3q1 n ajbjn Aj bj bj 1 Aq bq A p1 b p n n jp n jp n3 nn nn n b cn n q1 n n j p A j b j b j 1 n n Aq bq n n A p1 b p n n n cnn 3q1 n n b nA j n b j b j 1 n Aq n bq n A p1 n b p n jp r3 s c q1 b nM bj bj 1 bq b p bb j p c c M b p bq bq b p 2Mb p n 2Mb K Q3 R* n Since 0 was arbitrary, we conclude that ajbj is a Cauchy j0 n0 sequence of complex numbers. Therefore, it is convergent. A nice application of this result, gives us an easy to check criteria for convergence of series that are generated by sequences with alternating positive and negative terms. Theorem 4.5.40 (Alternating Series Test) Suppose that the sequence u j * 1 t j U satises the following conditions: 4.5. SERIES OF COMPLEX NUMBERS (i) sgn u j of (ii) u j 1 175 sgn u j 1 for each j + MC 0, where sgn denotes the sign n u j for every j and 0. (iii) lim u j j* 3 Then * 1 u j is convergent. Furthermore, if the sum is denoted by S, then j Sn n S n Sn 1 for each n where Sn * 0 is the sequence of nth partial sums. n The result is an immediate nconsequence of Theorem 4.5.39 it follows upon setn ting a j 1 j and b j nc j n. As an illustration of how a regrouping argument can get us to the conclusion, we offer the following proof for your reading pleasure. 0. Then u 2k 1 0 and Proof. Without loss in generality, we can take u 0 u 2k 0 for k 0 1 2 3 . Note that for each n + MC 0 S2n u 0 u1 u 2 u 3 u 2n2 u 2n1 u 2n which can be regrouped as S2n u0 u 1 u2 u 3 u 4 u 2n1 u 2n . The rst arrangement justies that S2n * 0 is monotonically increasing while the n second yields that S2n u 0 for each n. By Theorem 4.3.14, the sequence S2n * 0 n is convergent. For lim S2n S, we have that S2n n S for each n n* Since S2n1 S2n u 2n , S2n1 S2n for each n + M. On the other hand, S2n 1 S2n1 u 2n u 2n 1 S2n1 These inequalities combined with S2n S2 u 1 u 2 , yield that the sequence * is a monotonically decreasing sequence that is bounded below. Again, S2n1 n 1 by Theorem 4.3.14, S2n1 * 1 is convergent. From (iii), we deduce that S2n1 n S also. We have that S2n1 o S because S2n1 * 1 is decreasing. Pulling this n together, leads to the conclusion that Sn converges to S where S n Sk for k odd and S o Sk when k is even. Remark 4.5.41 Combining the Alternating Series Test with Remarkn4.5.5 leads to 3 the quick observation that the alternating harmonic series * 1 1 is conditionn n ally convergent. 176 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW 4.5.4 Discussing Convergence When asked to discuss the convergence of a given series, there is a system that we 3 should keep in mind. Given the series * 0 u n : n 1. Check whether or not lim u n 0. If not, claim divergence by the kth term n* test if yes, proceed to the next step. 3 3 2. Check for absolute convergence by testing * 0 u n . Since * 0 u n is a n n series having nonnegative terms, we have several tests of convergence at our disposalComparison, Limit Comparison, Ratio, and Rootin addition to the possibility of recognizing the given series as directly related to a geometric or a p-series. 3 Practice with the tests leads to a better ability to discern which test to use. If * 0 u n converges, by any of the our tests, then we conclude n 3 3 that * 0 u n converges absolutely and we are done. If * 0 u n 3 diverges n n by either the Ratio Test or the Root Test, then we conclude that * 0 u n n diverges and we are done. 3 3. If * 0 u n diverges by either the Comparison Test or the Limit Comparison n 3 Test , then test * 1 u n for conditional convergenceusing the Alternating n Series Test if it applies. If the series involves nonreal complex terms, try checking the corresponding series of real and imaginary parts. Excursion 4.5.42 Discuss the Convergence of each of the following: * ; 32n1 n1 1. n2 1 2. * ; 1n n ln n n1 en 4.5. SERIES OF COMPLEX NUMBERS 3. * ; 1n n n1 177 n 2 4. * ; n1 1 1 :n ,: 1 5. * ; cosn: n1 n2 ***The ratio test leads to the divergence of the rst one. The second one is absolutely convergent by the root test. The third one diverges due to failure to pass the kth term test. The behavior of the fourth one depends on :: it diverges for : 1 and converges for : 1 from the ratio test. Finally the last one converges by comparison.*** 4.5.5 Rearrangements of Series * * ; ; 11 Given any series a j and a function f : MC 0 MC 0, the series a f j * ; is a rearrangement of the original series. Given a series a j and a rearrange* ; ment a f j , the corresponding sequence of nth partial sums may be completely jo jo jo jo different. There is no reason to expect that they would have the same limit. The commutative law that works so well for nite sums tells us nothing about what may happen with innite series. It turns out that if the original series is absolutely convergent, then all rearrangements are convergent to the same limit. In the last section of Chapter 3 in our text, it is shown that the situation is shockingly different for 178 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW conditionally convergent real series. We will state the result that is proved on pages 76-77 or our text. Theorem 4.5.43 Let * ; a j be a real series that converges conditionally. Then for j0 any elements in the extended real number system such that * n : n ; n * ; there exists a rearrangement of the given series a f j such that j0 n ; a f j j0 n ; a f j j0 *, lim inf n* : and lim sup n* ;. * ; Theorem 4.5.44 Let a j be a series of complex numbers that converges abso* ; lutely. Then every rearrangement of a j converges and each rearrangement conj0 j0 verges to the same limit. 4.6 Problem Set D 1. Use the denition to prove each of the following claims. Your arguments must be well written and make use of appropriate approaches to proof. (a) lim n2 n* n 2 in 1 1 3n 2 i 0 n* 2n 3 3n 2 3 (c) lim n* 2n 1 2 3n 1 2ni (d) lim n* n3 1 3n (e) lim 3i n* 1 in (b) lim 3 2i 4.6. PROBLEM SET D 179 2. Find the limits, if they exist, of the following sequences in U2 . Show enough work to justify your conclusions. |t u} 1n cos n * (a) n n n1 u}* |t 2 3 3n 1 2n (b) 4n 1 n 2 2 n1 u}* |t n2 1 n 5 1 3n (c) 2n 2 1 2n n 1 |t u}* sin nn 1 (d) 2 n n n1 |t u} cos nH sin nH H2 * (e) n n n1 3. Suppose that xn * 1 converges to x in Euclidean k-space. Show that A n x n : n + M C x is closed. tu 2 sin H j n 3n 4 4. For j n + M, let f j n . Find the limit of the following 4 j 2 n 2 2 jn 1 sequence in U5 , showing enough work to carefully justify your conclusions: f 1 n f 2 n f 3 n f 4 n f 5 n* 1 . n 5. Find the limit superior and the limit inferior for each of the following sequences. Q nH R* (a) n cos 2 n1 * ! 1 cos nH ! 2 (b) ! 1n n 2 ! } nH nH * (c) sin 1 cos 4 2 n1 | t u }* 1 1n 1 1n1 (d) 2 3 n2 n1 1 2n n | n1 180 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW 6. If an * 0 is a bounded sequence of complex numbers and bn * 0 is a sen n quence of complex numbers that converges to 0, prove that lim an bn 0. n* 7. If an * 0 is a sequence of real numbers with the property that an an n 1 for each n + MC 0, prove that an * 0 converges. n 2n 8. If an * 0 is a monotonically increasing sequence such that an n 1 1 n 1 n for each n + MC 0, must an * 0 converge? Carefully justify your response. n an n 9. Discuss the convergence of each of the following. If the given series is convergence and it is possible to nd the sum, do so. (a) * ;1 T 3 n1 n * ; n1 (b) 1 n n 2 * ;1 (c) 2n n n1 (d) * ; 2n n1 3 n3 en (e) * ;n n1 10. Prove the Limit Comparison Test. Suppose that an * 0 and bn * 0 are such that an o 0, bn o 0 for each n n * * ; ; 1 n + MC 0, and lim an bn L 0. Then either an and bn both n* n1 n1 converge or both diverge. [Hint: For sufciently large n, justify that 11. Suppose that an o 0 for each n + MC 0. 1 an 3 L L.] 2 bn 2 4.6. PROBLEM SET D (a) If * ; an converges and bn n1 181 * * ; ; bT T ak , prove that bn bn kn n1 c 1 converges. * ; (b) If an diverges and Sn verges. n1 n * ; ; bT ak , prove that Sn k1 n1 1 Tc Sn di- 12. For each of the following use our tests for convergence to check for absolute convergence and, when needed, conditional convergence. (a) * ; n1 1n t 2n i 3n 5 4n 2n 1 H * n sin ; 2 (b) 2 n 1 n1 (c) (d) * ; rT n1 * ; n1 u 2n 2 s T 21 1 2n n4 1! t u2 * ; 1 n (e) cos H n 1 n n2 (f) * ; 1 n2 1n n (g) * ; n1 32n t in 3 1 4n 1n 2 1 ut 1 5 t n1 2i un 1n 1 2 1 t un 2 3 u 1 3 5 2n 1 p 13. Justify that is absolutely convergent for 1 2 4 6 2n n1 p 2, conditionally convergent for 0 p n 2, and divergent for p n 0. * ; 14. Let (2 be the collection of innite sequences xn * 1 n of reals such that * ; n1 2 xn 182 CHAPTER 4. SEQUENCES AND SERIESFIRST VIEW converges and dene d x y V * ; n1 xn yn 2 for each x xn * 1 y n yn * 1 + (2 . Show that (2 d is a metric space. n 15. A sequence x n * 1 of reals is bounded if and only if there is a number m n such that xn n m for each n + M. Let M denote the collection of all bounded sequences, and dened d x y sup xn yn . Show that M d n+M is a metric space. 16. Let B be the collection of all absolutely convergent series and dene d x y * ; xn yn . Show that B d is a metric space. n1

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

N.C. State - BUS - 420
N.C. State - BUS - 420
N.C. State - BUS - 420
N.C. State - BUS - 420
N.C. State - BUS - 420
Cornell - H ADM - 449
HA4449/6649 Participation Evaluation GuidelinesProfessor Lisa Klein Pearo CRITERIA1 UNACCEPTABLE PreparationRarely demonstrates basic preparation of class readings and assignmentsPOINTS3 GOOD 4 EXCELLENTConsistently demonstrates thorough prepa
Cornell - H ADM - 449
Professor Lisa Klein PearoContent Guidelines for IMC Project Proposal Due: September 18 HA4449/HA6649 Fall 2008The purpose of the proposal stage is to have your project idea approved and to get early feedback on the definition and scope of the pr
Cornell - H ADM - 449
HA 4449/HA6649 INTEGRATED MARKETING COMMUNICATIONS Fall 2008Instructor: Office: Phone: Dr. Lisa Klein Pearo 545C Statler Hall 255-9163 E-mail: lkp22@cornell.edu Office Hours: Wed. 12 -2 pm, and by appt.Teaching Assistants: Angeline Stuma James Cho
Cornell - H ADM - 449
HA4449/6649 Fall 2008 Professor Lisa Klein Pearo Case Write-Up Grading Criteria Cases will be evaluated according to the following criteria:1. 2. 3. 4. 5. 6. 7. 8.Clear identification of the issue(s) and their relative importance Demonstration of
Cornell - H ADM - 241
HA 241 FALL 2007 MARKETING PRINCIPLESTuesday and Thursday, 1:25 2:40 Statler Hall Room 165Professor: Office:Dr. Lisa Klein Pearo 545C Statler HallE-mail: lkp22@cornell.edu Phone: 255-9163 Office Hours: Live: Tues. 3:00 4:00 p.m. Virtual: Wed
Berkeley - ECON - c181
International Trade, Economics 181 Problem Set 3: Imperfect Competition Due Date: November 7, Beginning of Class 1. For each of the following examples, explain whether this is a case of external or internal economies of scale: (a) Most musical wind i
Berkeley - ECON - c181
Prof. Harrison, Econ 181, Fall 051Economics 181: International Trade Midterm Solutions1 Short Answer (20 points)Please give a full answer. If you need to indicate whether the answer is true or false, please explain your answer. You must give a
Berkeley - ECON - c181
Assignment #5 Due Date: Tuesday, December 5 Beginning of Class 1. A small country can import a good at a world price of 10 per unit. The domestic supply curve of the good is: S = 50 + 5P The demand curve is D = 400 10P In addition, each unit of prod
Berkeley - ECON - c181
NAMETAMidterm Exam Economics 181 PLEASE SHOW YOUR WORK! PUT YOUR NAME AND TAs NAME ON ALL PAGES 100 Points Total PART I. Short-Answer. (40 points). Please explain your work whenever possible. Each of the following 8 questions is worth 5 points ea
Berkeley - ECON - c181
Prof. Harrison, Econ 181, Fall 061Economics 181: International Trade Midterm SolutionsPlease answer all parts. Please show your work as much as possible.1Short Answer (40 points)Please give a full answer. If you need to indicate whether th
Berkeley - ECON - c181
Economics 181: International Trade Assignment #2 Due Thursday, September 28, beginning of class Professor Harrison, Fall 2006 1. Specific Factors and TradeFinland is capital abundant relative to potential trading partners in the rest of the world.
Berkeley - ECON - c181
NAME: TA: Midterm Economics 181 International Trade Fall 2005Please answer all parts. Please show your work as much as possible. Part I (20 points). Short Answer. Please give a full answer. If you need to indicate whether the answer is true or false
Berkeley - ECON - c181
Economics 181 International Trade Assignment #4 Due Date: Beginning of class November 21 1. The nation of Bermuda is small and assumed to be unable to affect world prices. It imports strawberries at the price of 10 dollars per box. The Domestic Suppl
Berkeley - ECON - c181
NAME Midterm Exam Economics 181 PLEASE SHOW YOUR WORK! PUT YOUR NAME AND TAs NAME ON ALL PAGES 100 Points Total PART I. Short-Answer. (40 points). Please explain your work whenever possible. Each of the following 8 questions is worth 5 points each. 1
Berkeley - ECON - c181
Economics 181 Assignment 1 Due Date: Thursday, September 14 Beginning of Class 1. We are given the following labor input requirements: Cloth UK USA (a) 2 hours 3 hours Wheat 6 hours 2 hoursWhat is the relative price of cloth in terms of wheat in th
SUNY Buffalo - PHY - 107
PHY107 SPRING 2007 EXAM 1Circle your instructors name:Prof. Jong Han Prof. Xuedong Hu Prof. Avto KharchilavaPRINT YOUR NAME LAST NAME FIRST NAMESIGN YOUR NAMESTUDENT ID NUMBER INSTRUCTIONS This is a closed book exam except that you
SUNY Buffalo - PHY - 107
Chapter 4(4 -1)Motion in Two and Three DimensionsIn this chapter we will continue to study the motion of objects without the restriction we put in chapter 2 to move along a straight line. Instead we will consider motion in a plane (two dimension
SUNY Buffalo - PHY - 107
Chapter 3VectorsIn Physics we have parameters that can be completely described by a number and are known as scalars .Temperature, and mass are such parameters Other physical parameters require additional information about direction and are known
SUNY Buffalo - PHY - 107
Chapter 1MeasurementIn this chapter we will explore the following concepts:1. Measurement of a physical parameter 2. Units, Systems of units 3. Basic Units in Mechanics 4. Changing units 5. Significant figures(1-1)In Physics we carry out expe
SUNY Buffalo - PHY - 107
Chapter 2Motion Along a Straight LineIn this chapter we will study kinematics i.e. how objects move along a straight line. The following parameters will be defined: Displacement Average velocity Average Speed Instantaneous velocity Average and inst
SUNY Buffalo - CHE - 108
Modern MaterialsDavid P. White University of North Carolina, Wilmington Chapter 12Copyright 1999, PRENTICE HALL Chapter 12 1Modern Materials Goal for modern chemistry and chemists: design materials with specific properties. Achieved by: modify
SUNY Buffalo - CHE - 108
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45.b a c a d d b d e b d e e b c c a b d b c e b b a a b b b b c a d a c c e d a
Iowa State - HIST - 201
Hollander2008History 201Midterm I Review Sheet Part One: Identification (40 points): Six (6) of the following terms will appear on the exam and you will have to choose four (4) and identify them (i.e. briefly explain who or what it is, when, wh
Iowa State - HIST - 201
History 201HollanderPower &amp; Civilization Characteristics of civilization: Cities Religious and political institutions Writing systems Increased material complexityExplanations for the rise of civilization: Technological: civilization follo
Cornell - COMM - 3010
Art of WooChapter 1 How Woo Works: The Four Steps Relationship based persuasion follows a distinctive, repeatable four-step process that you can master to achieve your influence goals. Step 1: Survey Your Situation -This step requires you to see you
American Academy of Art - HIST - 101
Project 1 2008 Biblootheek WERKBLADEN DEELOPGAVE 1 Groep: 1.1 INFORMATIEBRONNEN 12 OPSPOREN TIJDSCHRIFTARTIKEL Hier zijn minstens drie manieren van opsporen te bedenken: 1233 OPSPOREN CONFERENCE PAPER - Ja/Nee, locatie: Congresnaam: 4 KENNISCENTRUM
USC - HP - 400M
HP 400 BorovayMIDTERM #2 Study Guide 3/29/08The midterm will cover the material presented in PowerPoint lectures, textbook reading assignments, and student class presentations for Classes #12 [2/27/08] through #18 [3/26/08] A. Lectures and Reading
USC - HP - 400M
HP 400 Midterm #1 Study Guide 2008 Borovay From Lectures and Textbook Assignments for Classes # 1 and 2: 1. Be able to define the term culture as presented in the lecture on Introduction to Culture A set of guidelines which tells us how to view the w
USC - HP - 400M
o Hi everyoneIve have received a few messages from people wondering about the study guide questions about the film Sicko and the diabetes guest lecture. o For the Sicko section, my rationale is that everyone who saw the movie should at least be able
USC - HP - 400M
HP 400 Final Exam Study GuideSpring 2008 BorovayThe final exam covers the lectures, presentations, guest lectures, film, and textbook readings from after the second mini-midterm [Class 21] through Class 28, plus an overall perspective on global hea
USC - HP - 300
Relapse Prevention and the Abstinence Violation EffectTranstheoretical Model to Relapse PreventionPreparation Contemplation ActionPrecontemplation Lapse or RelapseMaintenanceWhat is a Lapse/Relapse?Example: Tim is an alcoholic. His alcohol
Villanova - THL - 1051
Michael Penna Martin Luther, born in 1483 and died in 1546, was a priest of the Augustinian order. He lived n a monastery in Wittenberg, Germany that was known for its strict religious life. Traveling to Rome, he was appalled at the sight of the life
Villanova - THL - 1051
Michael Penna Einhards Life of Charlemagne was written by Einhard, Charlemagnes primary biographer and main contributor of knowledge about Charlemagne. The document attempted to portray Charlemagne as a great national and religious hero. In fact, the
Villanova - THL - 1051
Michael Penna The Sack of Constantinople was written by Nicetas Choniates in 1204 as part of his work History that covered the years 1180-1207. At the age of nine Choniates was sent from his native town of Chonai in Phrygia to the city of Constantino
Villanova - THL - 1051
Michael Penna The author of Eusebiuss Conversion of Constantine is Eusebius himself. Eusebius was the bishop of Caesaria, located in present-day Syria1. He is the author of Ecclesiastical History, his own interpretation of the history of Christianity
Villanova - THL - 1051
Michael Penna The author of the Testimonium Flavium is Josephus. The document is part of the larger Antiquities, which is compendium of Jewish history. The Testimonium Flavianum is a section of Antiquities dealing with the historical person of Jesus.
Villanova - THL - 1051
Michael Penna The Apologia for the Second Crusade was written by Saint Bernard of Clairvaux, who lived from 1090-1153. He made his defense in the wake of the disastrous outcome of the crusades. St. Bernard was a member of the Cistercian Order. This g
Villanova - SPA - 1121
Villanova - THL - 1051
Michael Penna The Passion of Saints Perpetua and Felicity is an account of the martyrdom, or passion, of Revocatus, Felicity, Saturninus, Secundulus and Perpetua. Perpetua, the author of the document, was a new mother as well as new to Christianity d
Villanova - THL - 1051
Michael Penna The Book of Pastoral Rule was written by Pope Gregory, who lived from 540 to about 604. Before becoming the bishop of Rome Gregory was the prefect of Rome, serving as governor, chief of police and chief justice. Because of his previous
Villanova - THL - 1051
Michael Penna The document is a guideline, or rule, of the order of monastic life. It outlines how one should live in a monastic community. The document is divided into eight chapters, ranging from prayer to chastity to forgiveness. In each chapter t
Lehigh - IE - 215
IE 215 Solutions for Problems due Sep 10, 2008 (Chs 21, 22) 21.31 Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density = 2.9 g/cm3, and thermal diffusivity = 0.8 cm2/s. The following cutting conditions are used: cu
Lehigh - IE - 215
IE 215 Solutions for Problems due Sep 17, 2008 (Ch 23) 23.3 A series of turning tests were conducted using a cemented carbide tool, and flank wear data were collected. The feed was 0.010 in/rev and the depth was 0.125 in. At a speed of 350 ft/min, fl
Lehigh - IE - 215
IE 215 Solutions for Problems due Sep 3, 2008 (Ch 21) 21.6 The cutting force and thrust force have been measured in an orthogonal cutting operation to be 300 lb and 291 lb, respectively. The rake angle = 10, width of cut = 0.200 in, chip thickness be
Lehigh - IE - 220
IE220: Introduction to Operations Research Fall 2008 Professor Plik o Homework 2 Due Thursday, September 29 in class or online through BlackboardImportant Reminders Show your work. Answers such as (x1 ; x2 ) = (2; 6) are incomplete and will not be
Lehigh - IE - 220
IE220: Introduction to Operations Research Fall 2008 Prof. Imre Plik o Homework 1 Solutions 3.1-8Let x1 denote the number of units of special risk insurance and x2 the number of units for mortgages. The optimization problem is max 5x1 + 2x2 3x1 + 2x
Lehigh - IE - 220
IE220: Introduction to Operations Research Fall 2008 Professor Plik o Homework 1 Due Thursday, September 11 in class or online through BlackboardImportant Reminders Show your work. Answers such as (x1 ; x2 ) = (2; 6) are incomplete and will not be
Texas A&M - CHEM - 107
Lab #1: Conservation LawsSummary:In Lab. #1: Conservation Laws we experimented with distilled water and acetone to see if mass and volume where conserved when the tow substances were mixed together. First we added 5mL of water and repeated these sa
Kansas - POLS - 306
Final Study Guide for POLS 306 Spring 2008(The Final Exam will be multiple choice and short answer questions) Part I. Terms and Phrases The following terms are form the lecture and the text. You need to know the definitions from the book and the lec
Kansas - HIST - 129
Outside Readings: 1. education, military, agriculture, industry 2. Southern blacks moving north 3. wounded Knee Massacre/Ghost Dance 4. Buffalo 5. False. Soviets wanted two fronts 6. Capture the Rhine-Rhur Industrial region of Germany 7. Tarzan 8. Pe
Kansas - HIST - 121
Niki Grewal HIST 121 March 13, 2008 Midterm Exam 1) The late colonial period in Latin America was marked by much unrest within Spain and Portugal. Napoleon came and conquered Spain kicking Ferdinand VII out of power and formed a junta government with
Kansas - POLS - 170
Anarchy- absence of a supranational authority in a nation; self help states, realism- all against all, liberals-contracts, cooperation for survival BofP- intl system states provide for their own security by having equal power or more than others, pre
Kansas - POLS - 320
Question: Picktwoofthefollowingsubstantivepolicyissues(education, environment,stemcellresearch,gunpolicy,homelandsecurityorforeign policy).Compareandcontrastthepolicyprocesses(theoryandstages) involvedineachissuearea.Whattheorybestexplainsthedevelopm
Colorado - APPM - 2350
Colorado - APPM - 2350
Colorado - APPM - 2350