Chapter 4 Sequences and Series–First 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 n1 an n+M or a1 a2 a3 a4 This is the only time that we use the set bracket notation in a different con- text. The distinction is made in the way that the indexing is communicated . For : * : ¿ an ,the an n1 is the constant sequence that “lists the term in nitely often,” : : : : 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’forn going from one to in¿nity is the constant sequence that “lists the term : in¿nitely often,” : : : while the set consisting of ‘a sub n’forn 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 n1 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 1 n1 | }* | }* | r s}* n2 3n 5 n3 1 log n nH 1n and sin . 3 n 47 n1 n 1 n1 n 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 SERIES–FIRST 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. 2 4 8 16 32 64 1. 5 7 9 11 13 15 3 7 11 2. 1 9 81 729 5 9 13 ***An equation that works for (1) is 2n2n 31 while (2) needs a different for- mula for the odd terms and the even terms one pair that works is 2n 12n 11 for n even and 3n1 when n is odd.*** As part of the bigger picture, pattern recognition is important in areas of math- ematics 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 under- standing 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 de¿nition of convergence from your frosh calculus course, you might notice that the de¿nition 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 ¿ * De nition 4.1.1 Let pn n1 denote a sequence of elements of a metric space S d * and p0 be an element of S. The limit of pn n1 is p0 as n tends to (goes to or ap- proaches) in¿nity if and only if d e 1 + U F 0 " 2M MM + MF 1nn M " dpn p0 We write either pn p0 or lim pn p0. n* Remark 4.1.2 The description M M indicates that “limit of sequence proofs” require justi¿cation or speci¿cation 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 de¿ned by LxM inf j + ] : x n j z { 1 and is sometimes referred to as the ceiling function. Note, for example, that 2 1, L22M 2,andL5M 5. Compare this to the greatest integer function, which is sometimes referred to as the Àoor function. | }* 2 Example 4.1.3 The sequence has the limit 0 in U. We can take M 1 2, t u t nu n1 z { 1 3 700 M 200, and M 234. Of course, three examples 100 350 3 z { 2 does not a proof make. In general, for 0,letM .Thenn M implies that z { 2 o 2 n 0 n n n n 1 2 n2n which, by Proposition 1.2.9 (#7) and (#5), implies that and n n . n 2 n n Using the de¿nition 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 SERIES–FIRST 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 F complete the proof of the claim that converges to i in . n 1 n1 (a) Scratch work towards a proof.Becausei+ F,itsuf¿ces to show that 1 in lim i. Suppose 0 is given. Then n* n 1 n n n n n n T T n n n n n n n1 in n n1 in i n 1n n 1 i n 2 2 n in n n n n n 1 n 1 n 1 n 1 n T T 2 2 whenever n. So taking M will work. (b) A proof. For 0,letM .Thenn+ M and n M T 1 2 implies that n which is equivalent to . Because 2 T 4 T 2 n 1 n and 2 0, we also know that . Consequently, if n 3 n M , then n n n n n n T n n n n n n n1 in n n1 in i n 1 n n 1 i n 2 n in n n n n n 1 n 1 n 1 n 1 5 Since 0 was arbitrary, we conclude that v t t n n uuw n n n1 in n 1 0 " 2M M + MF 1n n M " n in n 1 1 in i.e., . Finally, i 0 1 + F and lim n* n 1 | 6 } * 1 in F i yields that converges to i in . n 1 n1 4.1. SEQUENCES AND SUBSEQUENCES IN METRIC SPACES 127 T T T 2 2 T 2 ***Acceptable responses are (1) ,(2) ,(3)n 1, (4) 2, (5) ,(6) n n 1 in lim i.*** n* n 1 ¿ * De nition 4.1.5 The sequence pn n1 of elements in a metric space S is said to 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 n | }* n 1 2 converges to 0 in U consequently, is convergent in Euclidean 1-space. n | }* n 1 2 b c On the other hand, is divergent in U x + U : x 0 d where d n n1 denotes the Euclidean metric on U,dx 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 n1 is a sequence of elements in a metric space S d . Then rK L s 1p1q p q + S F lim pn p F lim pn q " q p . 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 n1 be a sequence of elements in a metric space S d for 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 SERIES–FIRST VIEW 1 the p / q. Then d p q 0 and we let d p q. Because lim pn p and 2 n* 0, there exists a positive integer M1 such that n M1 " d pn p similarly, lim pn q yields the existence of a positive integer M2 such that n* . 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 d p q 2 3 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 M2 " d pn q ,(2)d pn q (3) 1 d p q,(4)p q.*** 2 ¿ * De nition 4.1.9 The sequence pn n1 of elements in a metric space S d is bounded if and only if d e 2M2x M 0 F x + S F 1nn + M " d x pnM .