57 Thus, there exists n2 >n1 such that ℓ− 1 2 <an2 <ℓ+ 1 2 . In this way, we can construct a strictly increasing sequence {nk} of positive integers such that ℓ− 1 k <ank <ℓ+ 1 k . Therefore, limk →∞ ank =ℓ. We now prove the converse. Suppose that (ii) is satisfied. Given anyε >0, there exists N∈N such that an <ℓ+ ε 2 andℓ−ε <ank <ℓ+ε for all n≥Nandk ≥N. Fix anym≥N. Then we have sm =sup{ak : k ≥m}≤ℓ+ ε 2 <ℓ+ε. 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 anyε >0, there exists N∈Nsuch that an >ℓ−ε for all n≥N, and there exists a subsequence {ank}of {an} such that lim k→∞ ank =ℓ. 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) lim supn →∞ 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 lim infn →∞ an =ℓ and {ank} is a subsequence of {an} with limk →∞ ank =ℓ′. Then ℓ′ ≥ℓ.
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