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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 that is not monotone. The sequence given by (−1)n  n converges to 0, and therefore is Cauchy, but not monotone. b) A with an unbounded . 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 → ∞ 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 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 , give a direct argument that (xn + yn) is Cauchy without using the Cauchy Criterion or the Algebraic Theorem.

Proof. Let  > 0 be given, and suppose (xn) and (yn) are Cauchy sequences. Then, there is an N1 ∈ N such that whenever m, n >  N1 then |xm − xn| < 2 . Similarly, there is an N2 ∈ N such that

1  whenever m, n > N2 then |ym − yn| < 2 . Let N = max(N1,N2). Then, if m, n > N,

|xm + ym − (xn + yn)| = |(xm − xn) + (ym − yn)|   ≤ |x − x | + |y − y | < + = . 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 |xn| ≤ M1, and there is an M2 > 0 with |yn| ≤ M2. There is an N1 ∈ N such that whenever m, n > N1  then |xm − xn| < 2M . Similarly, there is an N2 ∈ N such that 2  whenever m, n > N2 then |ym − yn| < . Hence, 2M1

|xmym − xnyn| = |xmym − xnym + xnym − xnyn|   ≤ |xm − xn||ym| + |ym − yn||xn| < |ym| + |xn| 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 = |an − bn|

Proof. Since the sequences are Cauchy, they are convergent. Thus, by the Algebraic Limit Theorem, an −bn → c for some 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, |an − bn − c| < c/2. However, this means that

c/2 < an − bn < 3c/2,

so an − bn > 0, and therefore |an − bn| = an − bn which has a limit of c.

Similarly, if c < 0, the same argument shows that |an − bn| → c.

2 Finally, if c = 0, we see that

||an − bn| − 0| = |an − bn| = |an − bn − 0|,

so the fact that an − bn is convergent leads us to |an − bn| 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.

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