15. a satisfies the Cauchy criterion for convergence of a series if, for any ε > 0, there is an no such that m Σακε for any m>n> no. k=n (a) Prove using only the definitions and results of this section: if k-1 ak con- verges, then ak satisfies the Cauchy criterion for convergence. (b) Use Theorem 2.6.4 on the convergence of a Cauchy sequence to prove the converse. Theorem 2.6.4. A sequence is Cauchy if and only if it is convergent. Proof. We show first that a convergent sequence is Cauchy. Let lim xn = a, and choose & > 0. From the definition of the limit of a sequence, there is an no so that if n≥ no then xn-al< ε/2. Now xn-xml=xn-a+a-xml≤xn-a|+|xm-al, and so if m,n≥ no, we have xn-xm| <ε/2+/2 = ε. Thus {x} is a Cauchy sequence. To verify the converse, suppose that the sequence {x} is Cauchy. First we show that this supposition implies that {x} is bounded. Because the sequence is Cauchy, there is an no such that if m, n ≥ no, then xn-xm| < 1. In particular, for all n ≥ no, we have xnxno <1, and therefore |xn| <1+|xo| (see Problem 5 of Section 1.3). Thus 1+xno is a bound for all but a finite number of the terms of the sequence, and if we define M = Max {x1, x2,...,xno-1,1+xnol}, we have |xn|M for all n. The sequence {x} is therefore bounded. Thus we can apply the Bolzano-Weierstrass Theorem: {x} must have a con- vergent subsequence, say {n}, with limk-> Xnk = a. We complete the proof of the theorem by showing that lim∞ Xn = a. Let & > 0. Because {x} is Cauchy, there is an no such that xn-xml<e/2 for m,n no. Because xnk → a, there is an n > no such that xn-al<e/2. Then for any n≥ no, xn-a=xnxnx + xnx - a≤ xnxn|+|xn-al<e/2+ε/2= ε. The sequence {x} converges to a. ☐