2.6: The Cauchy Criterion 2.6.2 Give an example of each of the following, or argue that such a request is impossible. a) A Cauchy sequence that is not monotone. The sequence given by (−1)n n converges to 0, and therefore is Cauchy, but not monotone. b) A Cauchy sequence with an unbounded subsequence. Since we know every Cauchy sequence is convergent, and every subsequence of a convergent sequence is convergent, this is impos- sible. c) A divergent monotone sequence, with a Cauchy subsequence. Since a bounded monotone sequence converges, if (an) is a diver- gent monotone sequence, an ! 1 or an ! −∞. If (ank ) is a subsequence, then the fact that nk ≥ k means that ank > ak. Since ak is increasing without bound, the subsequence must be also. Thus, this is impossible. d) An unbounded sequence containing a subsequence that is not Cauchy. Let ( n if n is odd. an = 1 n if n is even. Note that this sequence is unbounded, because for any natural number N, 2N + 1 > N, and a2N+1 = 2N + 1 > N. However, the 1 subsequence (a2n) is the sequence ( 2k ) which converges to 0. 2.6.3 a) If (xn) and (yn) are Cauchy sequences, give a direct argument that (xn + yn) is Cauchy without using the Cauchy Criterion or the Algebraic Limit Theorem. Proof. Let > 0 be given, and suppose (xn) and (yn) are Cauchy sequences. Then, there is an N1 2 N such that whenever m; n > N1 then jxm − xnj < 2 . Similarly, there is an N2 2 N such that 1 whenever m; n > N2 then jym − ynj < 2 . Let N = max(N1;N2). Then, if m; n > N, jxm + ym − (xn + yn)j = j(xm − xn) + (ym − yn)j ≤ jx − x j + jy − y j < + = . m n m n 2 2 Thus, (xn + yn) is Cauchy. b) Let > 0 be given, and suppose (xn) and (yn) are Cauchy se- quences. By Lemma 2.6.3, Cauchy sequences are bounded, so there is an M1 > 0 with jxnj ≤ M1, and there is an M2 > 0 with jynj ≤ M2. There is an N1 2 N such that whenever m; n > N1 then jxm − xnj < 2M . Similarly, there is an N2 2 N such that 2 whenever m; n > N2 then jym − ynj < . Hence, 2M1 jxmym − xnynj = jxmym − xnym + xnym − xnynj ≤ jxm − xnjjymj + jym − ynjjxnj < jymj + jxnj 2M2 2M1 ≤ + = . 2 2 Thus, (xnyn) is Cauchy. 2.6.4 Let (an) and (bn) be Cauchy sequences. Decide whether each of the following sequences is a Cauchy sequence. a) cn = jan − bnj Proof. Since the sequences are Cauchy, they are convergent. Thus, by the Algebraic Limit Theorem, an −bn ! c for some real number c. There are 3 cases to consider. Suppose c > 0. Then, if we let = c=2, there is an N such that if n > N, jan − bn − cj < c=2. However, this means that c=2 < an − bn < 3c=2; so an − bn > 0, and therefore jan − bnj = an − bn which has a limit of c. Similarly, if c < 0, the same argument shows that jan − bnj ! c. 2 Finally, if c = 0, we see that jjan − bnj − 0j = jan − bnj = jan − bn − 0j; so the fact that an − bn is convergent leads us to jan − bnj is convergent. n b) cn = (−1) an. n (−1)nn Suppose an = n+1 . Then (an) ! 1. However, n+1 does not converge, and so is not Cauchy. c) cn = [[an]]. Let ( 1 1 + n if n is even an = 1 1 − n if n is odd. Notice that ( 1 if n is even [[an]] = 0 if n is odd. This does not converge. 3.