Introduction to Mathematical Analysis I - Second Edition

39 Since { b n } converges to b , there is N 1 ∈ N such that | b n − b | < | b | 2 for n ≥ N 1 . It follows (using a triangle inequality) that, for such n , − | b | 2 < | b n |−| b | < | b | 2 and, hence, | b | 2 < | b n | . For each n ≥ N 1 , we have the following estimate 1 b n − 1 b = | b n − b | | b n || b | ≤ 2 | b n − b | b 2 . (2.2) Now let ε > 0. Since lim n → ∞ b n = b , there exists N 2 ∈ N such that | b n − b | < b 2 ε 2 for all n ≥ N 2 . Let N = max { N 1 , N 2 } . By ( 2.2 ) , one has 1 b n − 1 b ≤ 2 | b n − b | b 2 < ε for all n ≥ N . It follows that lim n → ∞ 1 b n = 1 b . Finally, we can apply part (c) and have lim n → ∞ a n b n = lim n → ∞ a n 1 b n = a b . The proof is now complete. Example 2.2.1 Consider the sequence { a n } given by a n = 3 n 2 − 2 n + 5 1 − 4 n + 7 n 2 . (2.3) Dividing numerator and denominator by n 2 , we can write a n = 3 − 2 / n + 5 / n 2 1 / n 2 − 4 / n + 7 (2.4) Therefore, by the limit theorems above, lim n → ∞ a n = lim n → ∞ 3 − 2 / n + 5 / n 2 1 / n 2 − 4 / n + 7 = lim n → ∞ 3 − lim n → ∞ 2 / n + lim n → ∞ 5 / n 2 lim n → ∞ 1 / n 2 − lim n → ∞ 4 / n + lim n → ∞ 7 = 3 7 . (2.5) Example 2.2.2 Let a n = n √ b , where b > 0. Consider the case where b > 1. In this case, a n > 1 for every n . By the binomial theorem , b = a n n = ( a n − 1 + 1 ) n ≥ 1 + n ( a n − 1 ) . This implies 0 < a n − 1 ≤ b − 1 n .