Proving Theorem: "If a Sequence Converts...

[JG89]

Homework Statement

I recently posted a question asking to prove the following theorem:

"If a sequence a_1 + a_2 + ... converges and if b_1, b_2, ... is a bounded monotonic sequence of numbers, then (a_1)(b_1) + (a_2)(b_2) + ... converges"

Here is a proof that I came across for it:

Let s_n denote the partial sums of \sum_{v=1}^n a_v, s the sum, and let \xi_n = s_n - s. Then \sum_{v=n}^m a_v b_v = \sum_{v=n}^m (\xi_v - \xi_{v-1}) b_v = \sum_{v=n}^m \xi_v(b_v - b_{v+1}) - \xi_{n-1} b_n + \xi_m b_{m+1}.

For every sufficiently large v, |\xi_v| < \epsilon, and

\sum_{v=n}^m a_v b_v < \epsilon \sum_{v=n}^m |b_v - b_{v+1}| + \epsilon |b_n| + \epsilon |b_{m+1}| < \epsilon |b_n - b_{m+1}| + \epsilon |b_n| + \epsilon |b_{m+1}|.

This is in turn less than 4B \epsilon, where B is a bound for |b_v|, and the series \sum_{v=1}^{\\infty} a_v b_v converges

Homework Equations

The Attempt at a Solution

I understand the proof and everything. I was wondering though, how did the writer of the proof know to rewrite the sum as this: \sum_{v=n}^m a_v b_v = \sum_{v=n}^m (\xi_v - \xi_{v-1}) b_v = \sum_{v=n}^m \xi_v(b_v - b_{v+1}) - \xi_{n-1} b_n + \xi_m b_{m+1} ?

It just seems so random, something that I never would've thought about. If you could, could you please explain the thought processes he went through to realize he had to rewrite the sum in that form?

Thanks

 

[Billy Bob]

That formula is called "summation by parts." It is the discrete analog of integration by parts.

 

[JG89]

Ah, thanks! The wiki article on it isn't loading properly for whatever reason. I'll wait until tomorrow to give it another look. If I can't find what I'm looking for on the formula, I'll be back!

Thanks