38 2.1 Convergence Definition 2.1.2 A sequence {an} is bounded above if the set {an : n ∈N} is bounded above. Similarly, the sequence {an}is bounded belowif 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 Definition 1.5.1 that the sequence {an}is bounded if and only if there is M∈Rsuch that |an| ≤Mfor 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∈Nsuch 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| ≤Mfor all n∈N. Therefore, {an}is bounded. □ Definition 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}anda2k+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. Nownk+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 aand let ε >0 be given. Then there exists Nsuch 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 aas k →∞. □
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