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 1−k| a2 −a1|. 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. ■ Example 2.4.1 Let an =lnn for all n∈N. Since 1< n+2 n+1 < n+1 n for all n∈Nand the natural logarithm is an increasing function, we have |an+2 −an+1| =|ln(n+2)−ln(n+1)| = ln n+2 n+1 =ln n+2 n+1 <ln n+1 n =|ln(n+1)−lnn| =|an+1 −an|. Therefore, the inequality in Definition 2.4.2 is satisfied 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 an = ncos(3n2 +2n+1) n+1 has a convergent subsequence.
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