57 Thus, there exists n2 > n1 such that 1 1 ℓ − <ℓ + . < an2 2 2 In this way, we can construct a strictly increasing sequence {nk} of positive integers such that 1 1 ℓ − <ℓ + . < ank k k Therefore, limk →∞ ank = ℓ. We now prove the converse. Suppose that (ii) is satisfed. Given any ε > 0, there exists N ∈ N such that ε an <ℓ + and ℓ − ε < ank <ℓ + ε 2 for all n ≥ N and k ≥ N. Fix any m ≥ N. Then we have ε sm = sup{ak : k ≥ m}≤ ℓ + <ℓ + ε. 2 By Lemma 2.1.5, nm ≥ m, so we also have sm = sup{ak : k ≥ m}≥ anm >ℓ − ε. Therefore, limm →∞ sm = limsupn →∞ an = ℓ. □ The next result can be proved in a similar way. Theorem 2.5.5 Let {an} be a sequence and let ℓ ∈ R. The following are equivalent: (i) liminfn →∞ an = ℓ. (ii) For any ε > 0, there exists N ∈ N such that an >ℓ − ε for all n ≥ N, and there exists a subsequence {ank } of {an} such that lim ank = ℓ. k→∞ The following corollary follows directly from Theorems 2.5.4 and 2.5.5. Corollary 2.5.6 Let {an} be a sequence and let ℓ ∈ R. Then the following are equivalent: (i) limn →∞ an = ℓ (ii) limsupn →∞ an = liminfn →∞ an = ℓ. Corollary 2.5.7 Let {an} be a sequence and let ℓ,ℓ ′ be real numbers. (i) Suppose limsupn →∞ an = ℓ and {ank } is a subsequence of {an} with limk →∞ ank = ℓ ′ . Then ℓ ′ ≤ ℓ. (ii) Suppose liminfn →∞ an = ℓ and {ank } is a subsequence of {an} with limk →∞ ank = ℓ ′ . Then ℓ ′ ≥ ℓ.
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