42 2.2 Limit Theorems Therefore, limn→∞(an +bn) =a+b. This proves (i). (ii) If k =0, then ka=0 and kan =0 for all n. The conclusion follows immediately. Suppose next that k̸ =0. Givenε >0, let N∈Nbe such that |an −a| <ε/|k| for n≥N. Then for n≥N, |kan −ka| =|k(an −a)| =|k||an −a| <|k| ε |k| =ε. It follows that limn→∞(kan) =kaas desired. This proves (ii). (iii) Since {an} is convergent, it follows from Theorem 2.1.4 that it is bounded. Thus, there exists M>0 such that |an| ≤Mfor all n∈N. For every n∈N, we have the following estimate: |anbn −ab| =|anbn −anb+anb−ab| ≤ |an||bn −b| +|b||an −a|. (2.1) Let ε >0. Since {an}converges to a, we may choose N1 ∈Nsuch that |an −a| < ε 2(|b| +1) for all n≥N1. Similarly, since {bn}converges to b, we may choose N2 ∈Nsuch that |bn −b| < ε 2M for all n≥N2. Let N=max{N1,N2}. Then, for n≥N, it follows from (2.1) that |anbn −ab| <M ε 2M +|b| ε 2(|b| +1) = ε 2 + | b| |b| +1 ε 2 < ε 2 + ε 2 =ε for all n≥N. Therefore, limn→∞anbn =ab. This proves (iii). (iv) We first show that lim n→∞ 1 bn = 1 b . Since {bn}converges to b, there is N1 ∈Nsuch that |bn −b| <| b| 2 for n≥N1. Using Corollary (1.4.4) it follows that, for suchn, − | b| 2 <|bn|−|b| <| b| 2 and, hence, | b| 2 <|bn|. For eachn≥N1, we have the following estimate 1 bn − 1 b =| bn −b| |bn||b| ≤ |bn −b| |b| 2 | b| = 2|bn −b| b2 . (2.2)
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