Introduction to Mathematical Analysis I - Second Edition

53 (a) Suppose limsup n → ∞ a n = ` and { a n k } is a subsequence of { a n } with lim k → ∞ a n k = ` 0 . Then ` 0 ≤ ` . (b) Suppose liminf n → ∞ a n = ` and { a n k } is a subsequence of { a n } with lim k → ∞ a n k = ` 0 . Then ` 0 ≥ ` . Proof: We prove only (a) because the proof of (b) is similar. By Theorem 2.5.4 and the definition of limits, for any ε > 0, there exists N ∈ N such that a n < ` + ε and ` 0 − ε < a n k < ` 0 + ε for all n ≥ N and k ≥ N . Since n N ≥ N , this implies ` 0 − ε < a n N < ` + ε . Thus, ` 0 < ` + 2 ε and, hence, ` 0 ≤ ` because ε is arbitrary. Remark 2.5.8 Let { a n } be a bounded sequence. Define A = { x ∈ R : there exists a subsequence { a n k } with lim a n k = x } . Each element of the set A called a subsequential limit of the sequence { a n } . It follows from Theorem 2.5.4 , Theorem 2.5.5 , and Corollary 2.5.7 that A 6 = /0 and limsup n → ∞ a n = max A and liminf n → ∞ a n = min A . Theorem 2.5.9 Suppose { a n } is a sequence such that a n > 0 for every n ∈ N and limsup n → ∞ a n + 1 a n = ` < 1 . Then lim n → ∞ a n = 0. Proof: Choose ε > 0 such that ` + ε < 1. Then there exists N ∈ N such that a n + 1 a n < ` + ε for all n ≥ N . Let q = ` + ε . Then 0 < q < 1. By induction, 0 < a n ≤ q n − N a N for all n ≥ N . Since lim n → ∞ q n − N a N = 0, one has lim n → ∞ a n = 0. By a similar method, we obtain the theorem below.