Math 430/530: Advanced Calculus I Chapter 1: Sequences Fall 2018
Sequences A sequence is just a list of real numbers a 1, a 2,... with a definite order. We write a = (a n ) n=1 for a sequence a with n th term a n. We can also view a sequence as the image of a function N R given by n a n. Very often, sequences will be given by a formula for the n th term, e.g. a n = 1/n or a n = ( 1) n. On other occasions, the first few terms of a sequence are given and we can infer the pattern. For example, the sequence 0, 1, 0, 1,... could also be written as a n = 1+( 1)n 2.
Convergence of sequences Informally, we want to define a sequence to be convergent if the terms of the sequence approach a particular number. Would these sequences converge: (i) a n = 0, (ii) a n = 1/n, (iii) a n = ( 1) n. Intuitively we want to define a, but this doesn t make sense because we only have a n for n N. How do we get around this? We formulate a definition based on the following four step strategy. Suppose we want a n to converge to L. (i) Let s give ourselves a target by considering a neighbourhood of L of radius ɛ for some small ɛ > 0. (ii) If we go far enough along the sequence, we should be able to always land in this neighbourhood. So there is some N N, such that... (iii) If n N. (iv) We conclude that a n L < ɛ.
Convergence of sequences To reiterate, the four steps are: (i) Target: given ɛ > 0 (ii) Qualifier: there exists N N (iii) Condition on the qualifier: such that if n N (iv) Conclusion/Hit the target: we have a n L < ɛ. A sequence (a n ) n=1 converges to a real number L if and only if for every ɛ > 0 there exists N N such that for all n N we have a n L < ɛ. Note that the smaller we take ɛ to be, the larger we typically need to take N to be.
s The sequence a n = 1/n converges to the real number 0. The sequence a n = 1 + 1/n 3 converges to 1. The sequence a n = ( 1) n does not converge. We can negate the definition of convergence as follows. The sequence (a n ) n=1 does not converge to L if there exists ɛ > 0, such that for every N N there exists n N such that a n L ɛ. In words, there is a definite positive quantity ɛ such that however far along the sequence we go, there is an element further along that is at least ɛ away from L.
Nbhds A set Q of real numbers is a neighbourhood (abbreviated to nbhd) of a real number x if and only if there exists ɛ > 0 such that (x ɛ, x + ɛ) Q. We can characterize convergence of sequences in terms on nbhds. Lemma A sequence (a n ) converges to L if and only if every nbhd of L contains all but finitely many terms of the sequence. Next, if a sequence converges, it can only converge to one limit. If (a n ) converges to A and (a n ) converges to B, then A = B.
Convergence and divergence A sequence (a n ) is called convergent if and only if there is a real number L so that a n converges to L. The value L is called the limit of the sequence. If the sequence is not convergent, it is called divergent. How can a sequence be divergent? If the sequence is unbounded, then it cannot be convergent so must be divergent. For example, a n = n gives a divergent sequence. Just because a sequence is bounded, doesn t mean it must converge. Sequences that oscillate cannot converge. For example, a n = ( 1) n is divergent. Other behaviour can occur. For example, if a 1 [0, 1] is irrational, and we define a n = 2a n 1 mod 1, then a n is a chaotic sequence which will not converge. (proof of this beyond the scope of this course)
Operations on sequences Suppose (a n ) converges to A and (b n ) converges to B. Then: (i) (a n + b n ) converges to A + B; (ii) (a n b n ) converges to AB; (iii) if in addition B 0, then (a n /b n ) converges to A/B; Find the limit of the sequence ( n 2 ) + 4n + 3 2n 2. + 6n 7 n=1 If a n 0 and (b n ) is a bounded sequence (not necessarily convergent), then (a n b n ) converges to 0.
Cauchy sequences A sequence (a n ) is called a Cauchy sequence if and only if for all ɛ > 0, there exists N N such that if m, n N, then a n a m < ɛ. For Cauchy sequences, we assume that all the terms beyond a certain point are close together. Every convergent sequence is Cauchy. Every Cauchy sequence is bounded.
Accumulation points Let S be any set of real numbers. We say that A R is an accumulation point of S if and only if every nbhd of A contains infinitely many points of S. Note that A does not have to be in S, e.g. the accumulation points of (0, 1) are [0, 1]. If (a n ) is a convergent sequence of distinct elements, then its limit L is an accumulation point of the set S = {a 1, a 2,...}. If S = {1/n : n N}, then 0 is the only accumulation point of S. Recalling the sequence a 1 [0, 1] irrational and a n = 2a n 1 mod 1, the set of accumulation points of {a 1, a 2,...} is all of [0, 1].
Bolzano-Weierstrass (Bolzano-Weierstrass ) Every bounded infinite set of real numbers has at least one accumulation point. For the set {(1 + ( 1) n )/n : n N}, 0 is the only accumulation points. A sequence viewed as a subset of R can have more than one accumulation point, for example {a n : a n = 1/n if n is even, a n = 1 + 1/n if n is odd } has accumulation points 0 and 1. Every Cauchy sequence in R is convergent.
Subsequences Let (a n ) n=1 be a sequence and (n k) k=1 be an increasing sequence of positive integers, i.e. n 1 < n 2 < n 3 <.... Then we say that (a nk ) k=1 is a subsequence of (a n ) n=1. The sequence (1/2k) k=1 is a subsequence of (1/n) n=1 with n k = 2k. The constant sequence ( 1) k=1 is a subsequence of (( 1) n ) n=1 with n k = 2k 1.
Two theorems on subsequences A sequence converges if and only if each of its subsequences converge to the same limit. Suppose that (a n ) is a bounded sequence. If all of its convergent subsequences have the same limit, then (a n ) is convergent. In the second theorem here, we do not initially assume that (a n ) is convergent.
Monotonic sequences A sequence (a n ) is called increasing if and only if a n a n+1 for all n N; it is called decreasing if and only if a n a n+1 for all n N; and it is called monotonic if and only if it is either increasing or decreasing. The sequence given by a n = 1/n is monotonic and decreasing. A monotonic sequence is convergent if and only if it is bounded. If c > 1, show that n c is monotonic, decreasing and bounded below; and hence it converges.