56 2.5 Limit Superior and Limit Inferior Proof: Suppose limsupn →∞ an = −∞. Then for any M ∈ R, there exists N ∈ N such that sn < M for all n ≥ N, where sn is defned in (2.8). In particular, sN = sup{ak : k ≥ N} < M, so ak ≤ sN < M for all k ≥ N. By the defnition, limn →∞ an = −∞. Let us now prove the converse. Suppose limn →∞ an = −∞. Then for any M ∈ R, there exists N ∈ N such that an < M − 1 < M for all n ≥ N. Take any n ≥ N. Since the set {ak : k ≥ n} is bounded above by M − 1, we get sn = sup{ak : k ≥ n}≤ M − 1 < M. It follows that limn →∞ sn = −∞. By the defnition, 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 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→∞ Proof: Suppose limsupn →∞ an = ℓ. Then limn →∞ sn = ℓ, where sn is defned in (2.8). For any ε > 0, there exists N ∈ N such 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 ∈ N such that ℓ − 1 < sN1 = sup{an : n ≥ N1} <ℓ + 1. Thus, there exists n1 ∈ N such that ℓ − 1 < an1 <ℓ + 1. 1 For ε = 2, there exists N2 ∈ N and N2 > n1 such that 1 1 ℓ − = sup{an : n ≥ N2} <ℓ + . < sN2 2 2
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