Convergence of Series with Monotonic and Bounded Sequences

[ForMyThunder]

Homework Statement

If \Sigmaa_n is a convergent series and {b_n} is a monotonic and bounded sequence, then \Sigma a_nb_nis a convergent series.

Homework Equations

The Attempt at a Solution

Since {b_n} is bounded, |b_n|<M for some M>0. Since \Sigmaa_n is a convergent series, we have that for every \epsilon>0, there is some N>0 such that |A_m-A_n|<\epsilon/M for all m>n>N. Thus, \sum_{k=n}^m a_kb_k < M \sum^{m}_{k=n} (A_k-A_k-1) < \epsilon.
And
<br /> \Sigma a_k b_k
is convergent.

Is this correct? If it is, then where did the monotonic behavior of {b_n} get put in the proof?

 

[Dick]

Your proof doesn't make any particular sense. But I think the type of case they are thinking of is an=(-1)^n/n and bn=(-1)^n. The series an is conditionally convergent, but an*bn is divergent.

 

[ForMyThunder]

I used the Cauchy criterion for convergence. Where does it not make sense?

{bn} is a monotone sequence by hypothesis so bn = (-1)^n wouldn't work for this.

 

[Dick]

It's works better than I thought it did now that I realize An is the nth partial sum of an. You should define your terms. But it still only works for an absolutely convergent series. I KNOW (-1)^n isn't monotone. You were asking why you need to assume bn to be monotone. I'm giving you a non-monotone example where your proof fails and asking you to try to figure out why it fails. Why do you need the monotone assumption?

 

[ForMyThunder]

I didn't think my proof was correct (since I don't see the use of the monotone assumption) so I was asking what was wrong with my proof; but if it was somehow correct, where is the monotone behavior hidden. I already know a proof for this but I was wondering why this particular one didn't work.

 

[Dick]

It's because of signs. (-1)<1. (-2)<(-1). But (-2)*(-1) is not less than 1*(-1). Do you see it now? You are assuming everything is positive.

 

[ForMyThunder]

Ahh. I see. Thanks. I've been obsessing over this for quite some time.

 

[JG89]

You are going to have to use Abel's summation by parts.

This is a pretty hard question. I remember being stuck on it last year.

 

[hunt_mat]

What about using the sandwich theorem?

 

[Tobias Funke]

I think you can use theorem 3.42 if you can find a way to define a new sequence c_n from b_n. I have that written in some old notes and if I didn't make a mistake the proof isn't long.