Introduction to Mathematical Analysis I - Second Edition

51 Theorem 2.5.2 If { a n } is not bounded above, then lim n → ∞ s n = ∞ , where { s n } is defined in ( 2.8 ) . Similarly, if { a n } is not bounded below, then lim n → ∞ t n = − ∞ , where { t n } is defined in ( 2.9 ) . Proof: Suppose { a n } is not bounded above. Then for any k ∈ N , the set { a i : i ≥ k } is also not bounded above. Thus, s k = sup { a i : i ≥ k } = ∞ for all k . Therefore, lim k → ∞ s k = ∞ . The proof for the second case is similar. Remark 2.5.3 By Theorem 2.5.2 , we see that if { a n } is not bounded above, then limsup n → ∞ a n = ∞ . Similarly, if { a n } is not bounded below, then liminf n → ∞ a n = − ∞ . Theorem 2.5.4 Let { a n } be a sequence and ` ∈ R . The following are equivalent: (a) limsup 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 = `. Proof: Suppose limsup n → ∞ a n = ` . Then lim n → ∞ s n = ` , where s n is defined as in ( 2.8 ) . For any ε > 0, there exists N ∈ N such that ` − ε < s n < ` + ε for all n ≥ N . This implies s N = sup { a n : n ≥ N } < ` + ε . Thus, a n < ` + ε for all n ≥ N . Moreover, for ε = 1, there exists N 1 ∈ N such that ` − 1 < s N 1 = sup { a n : n ≥ N 1 } < ` + 1 . Thus, there exists n 1 ∈ N such that ` − 1 < a n 1 < ` + 1 .