Introduction to Mathematical Analysis I - Second Edition

52 2.5 LIMIT SUPERIOR AND LIMIT INFERIOR For ε = 1 2 , there exists N 2 ∈ N and N 2 > n 1 such that ` − 1 2 < s N 2 = sup { a n : n ≥ N 2 } < ` + 1 2 . Thus, there exists n 2 > n 1 such that ` − 1 2 < a n 2 < ` + 1 2 . In this way, we can construct a strictly increasing sequence { n k } of positive integers such that ` − 1 k < a n k < ` + 1 k . Therefore, lim k → ∞ a n k = ` . We now prove the converse. Given any ε > 0, there exists N ∈ N such that a n < ` + ε and ` − ε < a n k < ` + ε for all n ≥ N and k ≥ N . Let any m ≥ N , we have s m = sup { a k : k ≥ m } ≤ ` + ε . By Lemma 2.1.8 , n m ≥ m , so we also have s m = sup { a k : k ≥ m } ≥ a n m > ` − ε . Therefore, lim m → ∞ s m = limsup m → ∞ a n = ` . The following result is proved in a similar way. Theorem 2.5.5 Let { a n } be a sequence and ` ∈ R . The following are equivalent: (a) liminf n → ∞ a n = ` . (b) For any ε > 0, there exists N ∈ N such that a n > ` − ε for all n ≥ N , and there exists a subsequence of { a n k } of { a n } such that lim k → ∞ a n k = `. The following corollary follows directly from Theorems 2.5.4 and 2.5.5 . Corollary 2.5.6 Let { a n } be a sequence. Then lim n → ∞ a n = ` if and only if limsup n → ∞ a n = liminf n → ∞ a n = `. Corollary 2.5.7 Let { a n } be a sequence.