I'm looking at a theorem that states if $\sum a_n$ and $\sum b_n$ are absolutely convergent then the Cauchy product $\sum c_n$ is absolutely convergent. Where $c_n = \sum_{j=0}^{n} a_j b_{n-j}$. I came up with this proof which is either 1) wrong because it doesn't use the fact that both series converge absolutely or 2) uses the fact that both series (particularly series $b_n$) converge absolutely without me noticing it. I have seen other similar posts such as this one but my question is about the following process specifically and looking at the previous posts I still can't tell if the problem is 1) or 2). I write this post in the hopes that someone will point out which one it is. Here is my proof:
Compute:
$$ \left|\sum_{n=0}^{M} |c_n| - \sum_{n=0}^{K} |c_n|\right|$$
Without loss of generality we can assume that $M \geq K$:
$$= \left|\sum_{n=K}^{M} |c_n| \right|=\sum_{n=K}^{M} |c_n|$$
Rewrite $c_n$:
$$=\sum_{n=K}^{M} |a_n B_{M-n}|$$
If $\sum b_n$ converges to $B$ and $B_M$ is the $M$'th partial sum of $b_n$ then:
$$=\sum_{n=K}^{M} |a_n (B_{M-n}-B+B)|$$
By the triangle inequality:
$$\leq \sum_{n=K}^{M} |a_n (B_{M-n}-B)| + \sum_{n=K}^{M} |a_n B|=\sum_{n=K}^{M} |a_n| |B_{M-n}-B| + \sum_{n=K}^{M} |a_n B|$$
Consider the set $\{|B_n - B| : n \in \mathbb{N}\}$ since $B_n$ converges then this set is bounded. Let $B_{MAX} = sup\{|B_n - B| : n \in \mathbb{N}\} $ then:
$$\leq B_{MAX} \sum_{n=K}^{M} |a_n| + |B| \sum_{n=K}^{M} |a_n| = (B_{MAX}+|B|) \sum_{n=K}^{M} |a_n|$$
Take $K$ large enough so $\sum_{n=K}^{M} |a_n| < \epsilon/{(B_{MAX}+|B|)}$ (which must exists since $a_n$ converges absolutely) to conclude:
$$\left |\sum_{n=0}^{M} |c_n| - \sum_{n=0}^{K} |c_n| \right|<\epsilon$$
Thus, $|c_n|$ is Cauchy and thus, $c_n$ converges absolutely.
QED?
