53 Proof: By induction, one has |an+1 − an|≤ k n−1|a2 − a1| for all n ∈ N. Thus, |an+p − an|≤|an+1 − an| + |an+2 − an+1| + ··· + |an+p − an+p −1| ≤ (kn−1 + kn + ··· + kn+p−2)|a2 − a1| ≤ kn−1(1+ k + k2 + ··· + kp−1)|a2 − a1| kn−1 ≤ |a2 − a1|. 1 − k for all n, p ∈ N. Since kn−1 → 0 as n → ∞ (independently of p), this implies {an} is a Cauchy sequence and, hence, it is convergent. □ The condition k < 1 in the previous theorem is crucial. Consider the following example. n+2 n+1 ■ Example 2.4.1 Let an = lnn for all n ∈ N. Since 1 < for all n ∈ N and the natural n+1 < n logarithm is an increasing function, we have n+ 2 n + 2 |an+2 − an+1| = |ln(n + 2) − ln(n + 1)| = ln = ln n+ 1 n + 1 n + 1 < ln = |ln(n+ 1) − lnn| = |an+1 − an|. n Therefore, the inequality in Defnition 2.4.2 is satisfed with k = 1, yet the sequence {lnn} does not converge since it is not bounded. Exercises 2.4.1 ▶ Determine which of the following are Cauchy sequences. (a) an =(−1) n . (b) an =(−1) n/n. (c) an = n/(n+ 1). (d) an =(cosn)/n. 2.4.2 Prove that the sequence ncos(3n2 + 2n + 1) an = n + 1 has a convergent subsequence.
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