Introduction to Mathematical Analysis I - Second Edition

147 Let N = max { 2 N 1 , 2 N 2 + 1 } . Then | a n − ` | < ε whenever n ≥ N . Therefore, lim n → ∞ a n = `. This problem is sometimes very helpful to show that a limit exists. For example, consider the sequence defined by x 1 = 1 / 2 , x n + 1 = 1 2 + x n for n ∈ N . We will see later that { x 2 n + 1 } and { x 2 n } both converge to √ 2 − 1, so we can conclude that { x n } converges to √ 2 − 1. (b) Use a similar method to the solution of part (a). Exercise 2.1.8 . Consider the case where ` > 0. By the definition of limit, we can find n 1 ∈ N such that | a n | > `/ 2 for all n ≥ n 1 . Given any ε > 0, we can find n 2 ∈ N such that | a n − ` | < ` ε 4 for all n ≥ n 2 . Choose n 0 = max { n 1 , n 2 } . For any n ≥ n 0 , one has a n + 1 a n − 1 = | a n − a n + 1 | | a n | ≤ | a n − ` | + | a n + 1 − ` | | a n | < ` ε 4 + ` ε 4 ` 2 = ε . Therefore, lim n → ∞ a n + 1 a n = 1. If ` < 0, consider the sequence {− a n } . The conclusion is no longer true if ` = 0. A counterexample is a n = λ n where λ ∈ ( 0 , 1 ) . SECTION 2.2 Exercise 2.2.3 . (a) The limit is calculated as follows: lim n → ∞ p n 2 + n − n = lim n → ∞ √ n 2 + n − n √ n 2 + n + n √ n 2 + n + n = lim n → ∞ n √ n 2 + n + n = lim n → ∞ n p n 2 ( 1 + 1 / n )+ n = lim n → ∞ 1 p 1 + 1 / n + 1 = 1 / 2 .