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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,

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

  • 0
    Where did this random picture come from?2012-09-25
  • 0
    If you're talking about the infant, it is maybe somepicture you use in other pages. Gravatars might sync in openIDs, I guess.2012-09-25
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    what you wrote is not cauchy product.2013-03-10

3 Answers 3