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Given $\sum a_n$ and $\sum b_n$ we define the Cauchy product to be $\sum_{k = 0}^{n}a_kb_{n-k}$. I need to prove that if both $\sum a_n$ and $\sum b_n$ are absolutely convergent then so is the Cauchy product.

Unless I'm missing something the proof seems trivial to me. Theorem 3.50 in the book "baby Rudin" states that given two convergent series at least one of which is absolutely convergent the Cauchy product will converge to the product of the limits. So using this why not just say the following?

Since $\sum_{n=0}^\infty |a_n|$ is convergent it is also absolutely convergent. Therefore, by the theorem 3.50,

$$\sum_{n=0}^\infty \sum_{k=0}^{n}|a_kb_{n-k}| = \sum_{n=0}^\infty |a_n| \sum_{n=0}^\infty |b_n|$$

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Where did this random picture come from? –  AAA Sep 25 '12 at 22:07
If you're talking about the infant, it is maybe somepicture you use in other pages. Gravatars might sync in openIDs, I guess. –  Pedro Tamaroff Sep 25 '12 at 22:42
what you wrote is not cauchy product. –  mez Mar 10 '13 at 17:44

2 Answers 2

Proof: Since $\sum_n |a_n|$ and $\sum_n |b_n|$ converges, let $\sum_n |a_n| < M$ and $\sum_n |b_n| < N$

$$\sum_{n = 0}^m |c_n| = \sum_{n = 0}^m|\sum_{k = 0}^n a_kb_{n-k}|\le\sum_{n = 0}^m\sum_{k = 0}^n |a_kb_{n-k}| $$ $$= |a_0b_0|+(|a_0b_1|+|a_1b_0|)+\cdots+(|a_0b_m|+|a_1b_{m-1}|+\cdots + |a_mb_0|)$$ $$ = \sum_{n = 0}^m |a_n| \sum_{k = 0}^{m-n} |b_k|< \sum_{n = 0}^m |a_n|N<NM$$ $\sum_n |c_n|$ is bounded above and monotone (since its sum of non-negative terms), it converges.
Thus Cauchy product of two absolutely convergent series converges absolutely.

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You did not write the Cauchy product correctly - where is the sum over $k$? When you fix this, you will need the triangle inequality and comparison test to finish the proof.

Yes, this is an easy exercise if you use Theorem 3.50. Arguably, it is a better exercise if you don't use Theorem 3.50 - a direct proof of the boundedness of partial sums is short and sweet.

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