1. Jun 24, 2009

### JG89

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

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

2. Relevant equations

3. 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

2. Jun 24, 2009

### Billy Bob

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

3. Jun 24, 2009

### 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