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Question about proof |
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| Jun24-09, 10:49 PM | #1 |
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Question about proof
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 [tex] \sum_{v=1}^n a_v [/tex], s the sum, and let [tex] \xi_n = s_n - s [/tex]. Then [tex] \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} [/tex]. For every sufficiently large v, [tex] |\xi_v| < \epsilon [/tex], and [tex] \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}| [/tex]. This is in turn less than [tex]4B \epsilon [/tex], where B is a bound for |b_v|, and the series [tex] \sum_{v=1}^{\\infty} a_v b_v [/tex] 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: [tex] \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} [/tex] ? 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 |
| Jun24-09, 11:09 PM | #2 |
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That formula is called "summation by parts." It is the discrete analog of integration by parts.
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| Jun24-09, 11:15 PM | #3 |
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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 |
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