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 definition of limits, for anyε >0, there exists N∈Nsuch that an < ℓ+ε andℓ′ −ε <ank < ℓ′ +ε for all n≥Nandk ≥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. Define 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 A̸=0/ and limsup n→∞ an =maxAand liminf n→∞ an =minA. Theorem 2.5.8 Suppose {an}is a sequence such that an >0 for every n∈Nand limsup n→∞ an+1 an =ℓ <1. Then limn→∞an =0. Proof: Choose ε >0 such that ℓ+ε <1. By Theorem 2.5.4, there exists N∈Nsuch that an+1 an < ℓ+ε for all n≥N. Let q=ℓ+ε. Then 0<q<1. By induction, 0<an ≤q n−NaN for all n≥N. Since limn→∞q n−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∈Nand liminf n→∞ an+1 an =ℓ >1. Then limn→∞an =∞. ■ Example 2.5.4 Given a real number α, define an = αn n! , n∈N. Whenα=0, it is obvious that limn→∞an =0. Suppose α>0. Then limsup n→∞ an+1 an =lim n→∞ α n+1 =0<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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