58 2.5 Limit Superior and Limit Inferior Proof: We prove only (i) because the proof of (ii) is similar. By Theorem 2.5.4 and the defnition of limits, for any ε > 0, there exists N ∈ N such that an <ℓ + ε and ℓ ′ − ε < ank <ℓ ′ + ε for all n ≥ N and k ≥ N. Since nN ≥ N, this implies ℓ ′ − ε < anN <ℓ + ε. Thus, ℓ ′ <ℓ + 2ε and, hence, ℓ ′ ≤ ℓ because ε > 0 is arbitrary. □ Remark 2.5.1 Let {an} be a bounded sequence. Defne A = {x ∈ R : there exists a subsequence {ank } with limank = x}. Each element of the set A is called a subsequential limit of the sequence {an}. It follows from Theorem 2.5.4, Theorem 2.5.5, and Corollary 2.5.7 that A ̸ = 0/ and limsupan = maxA and liminfan = minA. n→∞ n →∞ Theorem 2.5.8 Suppose {an} is a sequence such that an > 0 for every n ∈ N and limsup an+1 = ℓ< 1. n→∞ an Then limn →∞ an = 0. Proof: Choose ε > 0 such that ℓ + ε < 1. By Theorem 2.5.4, there exists N ∈ N such that an+1 <ℓ + ε for all n ≥ N. an Let q = ℓ + ε. Then 0 < q < 1. By induction, 0 < an ≤ q n−NaN for all n ≥ N. Since limn →∞ qn−NaN = 0, we have limn →∞ an = 0. □ By a similar method, we obtain the theorem below. Theorem 2.5.9 Suppose {an} is a sequence such that an > 0 for every n ∈ N and an+1 liminf = ℓ> 1. n→∞ an Then limn →∞ an = ∞. ■ Example 2.5.4 Given a real number α, defne αn an = , n ∈ N. n! When α = 0, it is obvious that limn →∞ an = 0. Suppose α > 0. Then an+1 α limsup = lim = 0 < 1. n→∞ an n→∞ n+ 1 Thus, limn →∞ an = 0. In the general case, we can also show that limn →∞ an = 0 by considering limn →∞ | an| and using Exercise 2.1.4.
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