Introduction to Mathematical Analysis I - Second Edition

49 Theorem 2.4.7 Every contractive sequence is convergent. Proof: By induction, one has | a n + 1 − a n | ≤ k n − 1 | a 2 − a 1 | for all n ∈ N . Thus, | a n + p − a n | ≤ | a n + 1 − a n | + | a n + 2 − a n + 1 | + · · · + | a n + p − a n + p − 1 | ≤ ( k n − 1 + k n + · · · + k n + p − 2 ) | a 2 − a 1 | ≤ k n − 1 ( 1 + k + k 2 + · · · + k p − 1 ) | a 2 − a 1 | ≤ k n − 1 1 − k | a 2 − a 1 | . for all n , p ∈ N . Since k n − 1 → 0 as n → ∞ (independently of p ), this implies { a n } is a Cauchy sequence and, hence, it is convergent. Example 2.4.1 The condition k < 1 in the previous theorem is crucial. Consider the following example. Let a n = ln n for all n ∈ N . Since 1 < n + 2 n + 1 < n + 1 n for all n ∈ N and the natural logarithm is an increasing function, we have | a n + 2 − a n + 1 | = | ln ( n + 2 ) − ln ( n + 1 ) | = ln n + 2 n + 1 = ln n + 2 n + 1 < ln n + 1 n = | ln ( n + 1 ) − ln n | = | a n + 1 − a n | . Therefore, the inequality in Definition 2.4.2 is satisfied with k = 1, yet the sequence { ln n } does not converge. Exercises 2.4.1 I Determine which of the following are Cauchy sequences. (a) a n = ( − 1 ) n . (b) a n = ( − 1 ) n / n . (c) a n = n / ( n + 1 ) . (d) a n = ( cos n ) / n . 2.4.2 Prove that the sequence a n = n cos ( 3 n 2 + 2 n + 1 ) n + 1 has a convergent subsequence.