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) 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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