56 2.5 Limit Superior and Limit Inferior Proof: Suppose limsupn →∞ an =−∞. Then for anyM∈R, there exists N∈Nsuch that sn <Mfor all n≥N, where sn is defined in (2.8). In particular, sN =sup{ak : k ≥N}<M, soak ≤sN <Mfor all k ≥N. By the definition, limn→∞an =−∞. Let us now prove the converse. Suppose limn→∞an =−∞. Then for any M∈R, there exists N∈Nsuch that an <M−1<Mfor all n≥N. Take any n≥N. Since the set {ak : k ≥n}is bounded above byM−1, we get sn =sup{ak : k ≥n} ≤M−1<M. It follows that limn→∞sn =−∞. By the definition, limsupn →∞ an =−∞. This completes the proof of (i). The proof of (ii) is similar. □ Theorem 2.5.4 Let {an}be a sequence and let ℓ ∈R. The following are equivalent: (i) limsupn →∞ 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. Proof: Suppose limsupn →∞ an =ℓ. Then limn→∞sn =ℓ, where sn is defined in (2.8). For any ε >0, there exists N∈Nsuch that ℓ−ε <sn < ℓ+ε for all n≥N. This implies sN =sup{an : n≥N}< ℓ+ε. Thus, an < ℓ+ε for all n≥N. Moreover, for ε =1, there exists N1 ∈Nsuch that ℓ−1<sN1 =sup{an : n≥N1}< ℓ+1. Thus, there exists n1 ∈Nsuch that ℓ−1<an1 < ℓ+1. For ε =1 2, there exists N2 ∈NandN2 >n1 such that ℓ− 1 2 <sN2 =sup{an : n≥N2}< ℓ+ 1 2 .
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