# Prove series converges

1. Jun 22, 2009

### JG89

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

If a_1 + a_2 + ... is an infinite series converging to A, and b1, b2, ... is an infinite sequence that is bounded and monotonic, prove that (a_1)(b_1) + (a_2)(b_2) + ... converges

2. Relevant equations

3. The attempt at a solution

I don't really know where to start...all I can say is that if a_1 + a_2 + ... converges, then a_n approaches 0 as n goes to infinity, and so (a_n)(b_n) also has a limit of 0, since b_n converges to some finite value.

2. Jun 22, 2009

### Office_Shredder

Staff Emeritus
You know that bn is bounded and monotonic. Have you tried using that? Given a general number b, you know
$$\sum a_n b$$
converges right? Now try to compare that to the sequence

$$\sum a_n b_n$$ choosing b such that |bn| < b for all n

3. Jun 22, 2009

### tiny-tim

Hi JG89!
Yes … concentrate on that value (call it b) …

then use deltas and epsilons.

4. Jun 23, 2009

### JG89

I have an intuitive idea of what's going on, but I'm having a hard time fleshing it out into an epsilon argument.

First off, since a_1 + a_2 + ... converges then b(a_1 + a_2 + ...) converges. I know that for large enough n, |(a_n)(b_n)| gets 'really' close to |(a_n)b|, since b is the limit of b_n. Since the b_n are monotonic, then as n increases the |(a_n)(b_n)| gets even closer to |(a_n)b|. I can in fact make this difference as small as I please, provided n is taken large enough and so it seems that after a certain n, the difference between (a_n)(b_n) and (a_n)b will become "negligible" and since (a_n)b is Cauchy, then (a_n)(b_n) is also Cauchy.

Last edited: Jun 23, 2009