38 2.1 Convergence Defnition 2.1.2 A sequence {an} is bounded above if the set {an : n ∈ N} is bounded above. Similarly, the sequence {an} is bounded below if the set {an : n ∈ N} is bounded below. We say that the sequence {an} is bounded if the set {an : n ∈ N} is bounded, that is, if it is both bounded above and bounded below. It follows from the observation after Defnition 1.5.1 that the sequence {an} is bounded if and only if there is M ∈ R such that |an|≤ M for all n ∈ N. Theorem 2.1.4 A convergent sequence is bounded. Proof: Suppose the sequence {an} converges to a. Then, for ε = 1, there exists N ∈ N such that |an − a| < 1 for all n ≥ N. Since |an|−|a| ≤ ||an|−|a|| ≤ |an − a|, this implies |an| < 1 + |a| for all n ≥ N. Set M = max{|a1|,...,|aN −1|,| a| + 1}. Then |an|≤ M for all n ∈ N. Therefore, {an} is bounded. □ Defnition 2.1.3 Let {an} ∞ n=1 be a sequence of real numbers. The sequence {bk} ∞ k=1 is called a subsequence of {an} ∞ n=1 if there exists a sequence of strictly increasing positive integers n1 < n2 < n3 < ··· , such that bk = ank for each k ∈ N. ■ Example 2.1.6 Consider the sequence {an} where an =(−1) n for n ∈ N. Then {bk} = {a2k} is a subsequence of {an} and a2k = 1 for all k (here nk = 2k for all k). Similarly, {ck} = {a2k+1} is also a subsequence of {an} and a2k+1 = −1 for all k (here nk = 2k + 1 for all k). Lemma 2.1.5 Let {nk}k be a sequence of positive integers with n1 < n2 < n3 < ···. Then nk ≥ k for all k ∈ N. Proof: We use mathematical induction. When k = 1, it is clear that n1 ≥ 1 since n1 is a positive integer. Assume nk ≥ k for some k. Now nk+1 > nk and, since nk and nk+1 are integers, this implies, nk+1 ≥ nk + 1. Therefore, nk+1 ≥ k + 1 by the inductive hypothesis. The conclusion now follows by the principle of mathematical induction (Theorem 1.3.1). □ Theorem 2.1.6 If a sequence {an} converges to a, then any subsequence {ank } of {an} also converges to a. Proof: Suppose {an} converges to a and let ε > 0 be given. Then there exists N such that |an − a| < ε for all n ≥ N. For any k ≥ N, since nk ≥ k, we also have |ank − a| < ε. Thus, {ank } converges to a as k → ∞. □
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