# Rudin convergence proof

1. Jul 21, 2010

### ForMyThunder

1. The problem statement, all variables and given/known data

If $$\Sigma$$an is a convergent series and {bn} is a monotonic and bounded sequence, then $$\Sigma$$ anbnis a convergent series.

2. Relevant equations

3. The attempt at a solution

Since {bn} is bounded, |bn|<M for some M>0. Since $$\Sigma$$an is a convergent series, we have that for every $$\epsilon$$>0, there is some N>0 such that |Am-An|<$$\epsilon$$/M for all m>n>N. Thus, $$\sum_{k=n}^m$$ akbk < M $$\sum^{m}_{k=n}$$ (Ak-Ak-1) < $$\epsilon$$.
And
$$\Sigma$$ ak bk
is convergent.

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

2. Jul 21, 2010

### 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.

3. Jul 21, 2010

### 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.

4. Jul 21, 2010

### 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?

5. Jul 21, 2010

### 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.

6. Jul 21, 2010

### 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.

7. Jul 21, 2010

### ForMyThunder

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

8. Jul 22, 2010

### 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.

9. Jul 22, 2010

### hunt_mat

What about using the sandwich theorem?

10. Jul 22, 2010

### 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.