Solutions to Practice Problems

Exercise 8.8(a) Show that if {an}n=1 is Cauchy then {a2n}n=1 is also Cauchy.(b) Give an example of a Cauchy sequence {a2n}n=1 such that {an}n=1 is notCauchy.

Solution.(a) Since {an}n=1 is Cauchy, it is convergent. Since the product of twoconvergent sequences is convergent the sequence {a2n}n=1 is convergent andtherefore is Cauchy.(b) Let an = (1)n for all n N. The sequence {an}n=1 is not Cauchy sinceit is divergent. However, the sequence {a2n}n=1 = {1, 1, } converges to 1so it is Cauchy

Exercise 8.9Let {an}n=1 be a Cauchy sequence such that an is an integer for all n N.Show that there is a positive integer N such that an = C for all n N,where C is a constant.

Solution.Let = 1

2. Since {an}n=1 is Cauchy, there is a positive integer N such that if

m,n N we have |am an| < 12 . But am an is an integer so we must havean = aN for all n N

Exercise 8.10Let {an}n=1 be a sequence that satisfies

|an+2 an+1| < c2|an+1 an| for all n N

where 0 < c < 1.(a) Show that |an+1 an| < cn|a2 a1| for all n 2.(b) Show that {an}n=1 is a Cauchy sequence.

Solution.(a) See Exercise 1.10.

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(b) Let > 0 be given. Since limn cn = 0 we can find a positive integer

N such that if n N then |c|n < (1 c). Thus, for n > m N we have

|an am| |am+1 am|+ |am+2 am+1|+ + |an an1| 0 be given. There exist positive integers N1 and N2 such that ifn,m N1 and n,m N2 we have |an am| < 2 and |bn bm| 0 be given. Let N be a positive integer to be chosen. Suppose thatn,m N. We have

|an am| = n+ 32n+ 1 m+ 32m+ 1

= 3 |m n|(2n+ 1)(2m+ 1) 2m+ 2n

(2n+ 1)(2m+ 1)=

(2n+ 1) + (2m+ 1) 2(2n+ 1)(2m+ 1)

=1

2m+ 1+

1

2n+ 1 2

(2n+ 1)(2m+ 1)

12m+ 1

+1

2n+ 1

22N + 1

Choose N so that 22N+1

< . That is N > 22. In this case,

|an am| <

for all n,m N. That is, { n+32n+1}n=1 is Cauchy

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Exercise 8.17Consider a sequence defined recursively by a1 = 1 and an+1 = an + (1)nn3for all n N. Show that such a sequence is not a Cauchy sequence. Doesthis sequence converge?

Solution.We will show that there is an > 0 such that for all N N there exist m andn such that m,n N but |am an| . Note that |an+1 an| = n3 1. Let = 1. Let N N. Choose m = N + 1 and n = N. In this case, |am an| =N3 1 = . Hence, the given sequence is not a Cauchy sequence. Since everyconvergent sequence must be Cauchy, the given sequence is divergent

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