Introduction to Mathematical Analysis I - Second Edition

35 Theorem 2.1.7 A convergent sequence is bounded. Proof: Suppose the sequence { a n } converges to a . Then, for ε = 1, there exists N ∈ N such that | a n − a | < 1 for all n ≥ N . Since | a n | − | a | ≤ || a n | − | a || ≤ | a n − a | , this implies | a n | < 1 + | a | for all n ≥ N . Set M = max {| a 1 | , . . . , | a N − 1 | , | a | + 1 } . Then | a n | ≤ M for all n ∈ N . Therefore, { a n } is bounded. Definition 2.1.3 Let { a n } ∞ n = 1 be a sequence of real numbers. The sequence { b n } ∞ n = 1 is called a subsequence of { a n } ∞ n = 1 if there exists a sequence of increasing positive integers n 1 < n 2 < n 3 < · · · , such that b k = a n k for each k ∈ N . Example 2.1.6 Consider the sequence a n = ( − 1 ) n for n ∈ N . Then { a 2 k } is a subsequence of { a n } and a 2 k = 1 for all k (here n k = 2 k for all k ). Similarly, { a 2 k + 1 } is also a subsequence of { a n } and a 2 k + 1 = − 1 for all k (here n k = 2 k + 1 for all k ). Lemma 2.1.8 Let { n k } k be a sequence of positive integers with n 1 < n 2 < n 3 < · · · Then n k ≥ k for all k ∈ N . Proof: We use mathematical induction. When k = 1, it is clear that n 1 ≥ 1 since n 1 is a positive integer. Assume n k ≥ k for some k . Now n k + 1 > n k and, since n k and n k + 1 are integers, this implies, n k + 1 ≥ n k + 1. Therefore, n k + 1 ≥ k + 1 by the inductive hypothesis. The conclusion now follows by the principle of mathematical induction . Theorem 2.1.9 If a sequence { a n } converges to a , then any subsequence { a n k } of { a n } also con- verges to a . Proof: Suppose { a n } converges to a and let ε > 0 be given. Then there exists N such that | a n − a | < ε for all n ≥ N . For any k ≥ N , since n k ≥ k , we also have | a n k − a | < ε . Thus, { a n k } converges to a as k → ∞ . Example 2.1.7 Let a n = ( − 1 ) n for n ∈ N . Then the sequence { a n } is divergent. Indeed, suppose by contradiction that lim n → ∞ a n = `.