Q:

Is there a simple proof that $(\Sigma a_n)(\Sigma b_n)=\Sigma (a_0b_n + \ldots + a_nb_0)$ whenever the series converge?

That is, if the power series $\Sigma a_n$, $\Sigma b_n$ and $\Sigma c_n$ converge to $A, B$ and $C$ respectively, where $c_n = a_0b_n + \ldots + a_nb_0$, then, $$AB=C$$
Rudin's proof, for instance, makes use of Merten's Theorem and the continuity of power series (in fact, he mentions the theorem in chapter 3, but doesn't prove it until chapter 8 due to lack of tools to establish the result). I was just wondering if there was a more direct proof.

A:

Replacing Abel summability by Cesàro summability, we can still prove an analogous statement.
Indeed, write $A_n = \sum_{k=0}^{n} a_k$, $B_n = \sum_{k=0}^{n} b_k$, and $C_n = \sum_{k=0}^{n} c_k$ for the partial sums. In view of Stolz–Cesàro theorem, it suffices to prove the following claim.

Proposition. Suppose that $A_n \to A$ and $B_n \to B$. Then
  $$ \lim_{n\to\infty} \frac{\sum_{k=0}^{n} C_k}{n+1} = AB. $$

Proof. By counting how many the term $a_i b_j$ appears in both sides, it is not hard to check that
$$ \sum_{k=0}^{n} C_k = \sum_{i + j \leq n} (n + 1 - i - j) a_i b_j = \sum_{k=0}^{n} A_k B_{n-k}. $$
Now for each $\epsilon > 0$, use the assumption to pick $N$ such that both $|A_n - A| < \epsilon$ and $|B_n - B| < \epsilon$ for $n \geq N$. Then with $M_A := \sup_n |A_n|$ and $M_B := \sup_n |B_n|$, we get
$$ \left| \frac{\sum_{k=0}^{n} C_k}{n+1} - AB \right|
= \left| \frac{\sum_{k=0}^{n} A_k B_{n-k}}{n+1} - AB \right|
\leq \frac{4NM_A M_B}{n+1} + (M_A + M_B) \epsilon, $$
and so, letting limsup as $n\to\infty$ gives
$$ \limsup_{n\to\infty} \left| \frac{\sum_{k=0}^{n} C_k}{n+1} - AB \right| \leq (M_A + M_B) \epsilon. $$
Since $\epsilon > 0$ is arbitrary, letting $\epsilon \downarrow 0$ proves the desired result. ////

