Proving Convergence of Series with Bounded Monotonic Sequences

[JG89]

Homework Statement

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

Homework Equations

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.

 

[Office_Shredder]

You know that b_n 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 |b_n| < b for all n

 

[tiny-tim]

Hi JG89!

… since b_n converges to some finite value.

Yes … concentrate on that value (call it b) …

then use deltas and epsilons.

 

[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.