Cauchy product on topological rings Let $R$ be any commutative Hausdorff topological ring. I am looking for a preferably general condition on sequences $(x_n)_{n \in \mathbb{N}}$, $(y_n)_{n \in \mathbb{N}}$ such that the equation $$ \left(\sum_{n=0}^\infty x_n \right)\left(\sum_{n=0}^\infty y_n \right) = \sum_{n=0}^\infty \sum_{k=0}^n x_k y_{n-k} $$
is valid. Is it possible to be general enough to cover the cauchy product of absolutely convergent series of real numbers as well as the ordinary product in any topological ring of formal power series (considered as the $(X)$-adic completion of a polynomial ring)?
Thank you in advance!
 A: $\sum_{n=0}^{\infty} x_n$ is the limit of the sequence $k \mapsto \sum_{n=0}^{k} x_n$ (if it exists, which I assume), similarily for $y$, hence $\left(\sum_{n=0}^{\infty} x_n\right) \cdot \left(\sum_{n=0}^{\infty} y_n\right)$ is the limit of the sequence $C : k \mapsto \left(\sum_{n=0}^{k} x_n \right) \cdot \left(\sum_{n=0}^{k} y_n\right)=\sum_{n=0}^{2k} \left(\sum_{p,q \leq k, p+q=n} x_p y_q\right)$. We want to compare this with the limit of the sequence $D : k \mapsto \sum_{n=0}^{k} \sum_{p+q=n} x_p y_q$. We have 
$C(k)-D(2k)=\sum_{n=0}^{2k} \sum_{p+q=n, p>k \vee q>k} x_p y_q=\sum_{n=0}^{2k} \sum_{k<p \leq n} x_p y_{n-p} +\sum_{n=0}^{2k} \sum_{k<q \leq n} x_{n-q} y_q$.
Now assume that $R$ has the $I$-adic topology for some ideal $I$. Let $m \in \mathbb{N}$. Since $x_p \to 0$ for $p \to \infty$, we have $x_p \in I^m$ for large $p$. Similarily, we have $y_p \in I^m$ for large $p$. From the above formula we see that $C(k) - D(2k) \in I^m$ for large $k$. This proves that $C(k) - D(2k) \to 0$ for $k \to \infty$. Similarily one proves that $C(k) - D(2k+1) \to 0$ (using $\sum_{n=0}^{k+1} y_n$). Thus, $D(2k)$ and $D(2k+1)$ have the same limit as $C$. It follows that $D$ converges, and has the same limit as $C$.
Summary: In an adic topological ring, if $\sum_n x_n$ and $\sum_n y_n$ converge, then also their Cauchy product converges and the formula holds.
Fun remark: In a cocomplete $\otimes$-category, the formula
$$\bigoplus_{n=0}^{\infty} X_n \otimes \bigoplus_{n=0}^{\infty} Y_n = \bigoplus_{n=0}^{\infty} \bigoplus_{p+q=n} X_p \otimes Y_q$$
holds without any conditions. I love these $\otimes$-categories ...
