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I know there are a couple of proofs of this result on this website, but I would like to verify if my proof is correct, as it is slightly different.

Let $\sum\limits_{n=0}^\infty a_n$ and $\sum\limits_{n=0}^\infty b_n$ be two absolutely convergent series of complex numbers, let $c_n = \sum\limits_{k=0}^n a_k b_{n-k}$. Prove that $\sum\limits_{n=0}^\infty c_n$ is absolutely convergent and that $$\sum\limits_{n=0}^\infty c_n = \left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right) \text{ (*) }$$

Proof:

(I'm not sure why it is asked to prove a definition, because (*) is a Cauchy product of infinite series, by definition).

Let $\sum\limits_{n=0}^\infty a_n = L$ and $\sum\limits_{n=0}^\infty b_n=M$. By definition of the Cauchy product, $\left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right)=\sum\limits_{k=0}^\infty\sum\limits_{l=0}^k a_l b_{k-l}=\sum\limits_{k=0}^\infty c_k$.

Now, $\left\|\left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right)\right\|\le \sum\limits_{n=0}^\infty \|a_n\|\sum\limits_{n=0}^\infty \|b_n\|=LM<\infty$, so $\sum\limits_{n=0}^\infty c_n$ converges absolutely.

If we didn't use the Cauchy product, then

$\sum\limits_{n=0}^N \|c_n\|=\sum\limits_{n=0}^N \|\sum\limits_{k=0}^n a_k b_{n-k} \|\le \sum\limits_{n=0}^N \sum\limits_{k=0}^n \|a_k b_{n-k} \|=\|a_0b_0\|+\|a_0b_1\|+...+\|a_Nb_0\|+...+\|a_Nb_N\|=\|a_0\|\sum\limits_{n=0}^N \| b_n\|+...+\|a_N\|\sum\limits_{n=0}^N\|b_n\|=(\|a_0\|+...+\|a_N\|)\sum\limits_{n=0}^N\|b_n\|=\sum\limits_{n=0}^N\|a_n\|\sum\limits_{n=0}^N\|b_n\|$. Now $\lim\limits_{N\to\infty}\sum\limits_{n=0}^N\|a_n\|\sum\limits_{n=0}^N\|b_n\|=LM$.

Do you think the above is correct? And which approach is better - the first or the second? It seems to me that the first is better since it is shorter.

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    How do you get the first inequality on your first method? – 2017-02-16
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    $\left\|\left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right)\right\|= \left\| \sum\limits_{n=0}^\infty a_n\right\|\left\|\sum\limits_{n=0}^\infty b_n\right\|\le \sum\limits_{n=0}^\infty \|a_n\|\sum\limits_{n=0}^\infty \|b_n\|$ – 2017-02-16
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    That assumes what you want to proof, one line before you used that notation to denote the Cauchy product, and the next line you assume it is equal to that product. – 2017-02-16
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    I'm just using the definition of the Cauchy product. – 2017-02-16
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    No, read what you just wrote, that's the regular product. Otherwise the first equality would be the definition of Cauchy's product, and you wouldn't be able to reach the second equality as easy. – 2017-02-16
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    https://en.wikipedia.org/wiki/Cauchy_product#Cauchy_product_of_two_infinite_series – 2017-02-16
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    I am not explaining myself properly. It is the notation you used that lead to this confusion, try denoting the Cauchy product by $*$ instead of $\cdot$ and you will see it more clearly. In your proof you assumed that $*=\cdot$, which is exactly what you want to show (when the series converge absolutely). – 2017-02-16
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    Do you think that my second method has in itself a hint on how to prove that $* = \cdot$? Actually I thought that the Cauchy product is the same as regular product, by definition. As it looks like regular multiplication of two infinite expressions like $(a_0+a_1+...)(b_0+b_1+...)$. – 2017-02-16
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    It is the same as the regular product for finite sums, but in general, if both series converge, but do not converge absolutely, it is not true that is the same. The second method shows that if both series absolutely converge (if I recall correctly, it is enough if only one of them does), then so does the Cauchy product, from there is easy to show that, in this case, $\cdot=*$. – 2017-02-16

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