$A = \sum_{n=0}^\infty a_n$ and $b_n \to B$ implies $\sum_{k=0}^n a_k b_{n-k} \to AB$

In electronics, we call such a sum : the numerical convolution of two signals. And like you've remarked, it is a Cauchy product.

This theorem was proved by Franz Mertens. What's missing from the above proof, is supposing that at least one of the series has to be absolutely convergent. Otherwise, the estimation of the first part of the second sum, isn't justified. A counterexample is given in the Wikipedia link.

As it turns out, the statement is equivalent to Merten's theorem about the Cauchy product of two series (with the first series being absolutely convergent).

We define $$ \begin{align} \beta_0 &= b_0 \\ \beta_n &= b_n - b_{n-1} \text{ for } n \ge 1 \, . \end{align} $$ and $\gamma_n = \sum_{k=0}^n a_k \beta_{n-k}$.

Then $b_n = \beta_0 + \ldots + \beta_n$ and $B = \sum_{n=0}^\infty \beta_n$. A simple calculation shows that $$ \sum_{k=0}^n a_k b_{n-k} = \sum_{k=0}^n \gamma_k \, . $$

Therefore $\lim_{n \to \infty} \sum_{k=0}^n a_k b_{n-k} = A B $ is equivalent to $ \sum_{n=0}^\infty \gamma_n = A B$, and that is precisely he contents of Merten's theorem.