I was going to ask this question, but I think I figured it out, so I thought I'd post my answer:
In this question of mine, a user's answer makes the following claim:
Suppose $r_n$ and $s_n$ are sequences of real numbers, converging to $r$ and $s$, respectively. Then we have $$ \frac{\sum_{j+k = M}r_js_k}{M+1} \xrightarrow{M\to\infty} rs. $$ How can we prove this?
Fix $\epsilon >0$. Choose $N_{\epsilon}$ such that $|r_is_j-rs|<\epsilon$ if $i,j\geq N_{\epsilon}$. For this I claim it is sufficient to choose $N_{\epsilon}$ such that
$$
|r_i-r|< \min \left( \dfrac{\epsilon}{2(|s|+1)},1\right) \quad \text{and}\quad |s_j-s|< \frac{\epsilon}{2(|r|+1)}.
$$
We want to show that
$$
\frac{\sum_{j+k = M}r_js_k}{M+1} \tag{*}
$$
converges to $rs$. Suppose $M\geq 2N_{\epsilon}$. We can break the sum in (*) into three parts:
$$
(M+1)^{-1}(\underbrace{r_0s_{M} + \dotsb + r_{N_{\epsilon}-1}s_{M-(N_{\epsilon}-1)}}_{(1)} + \underbrace{r_{N_\epsilon}s_{M-N_{\epsilon}} + \dotsb + r_{M-N_{\epsilon}}s_{N_{\epsilon}}}_{(2)} + \underbrace { \dotsb + r_{M}s_0}_{(3)}).
$$
I claim
$$
\text{(1) and (3) are bounded as $M\to \infty$.}\tag{**}
$$
For example, for (1) we have
$$
|r_0s_{M} + \dotsb + r_{N_{\epsilon}-1}s_{M-(N_{\epsilon}-1)}|\leq N_{\epsilon}(|s|+\epsilon)\cdot \max\limits_{0\leq i \leq N_{\epsilon}} |r_i|.
$$
Then ( * *) implies that (1) and (3) are negligible when compared with the denominator $M+1$. For (2) we have
$$
(M-2N_{\epsilon})(rs - \epsilon)\leq r_{N_\epsilon}s_{M-N_{\epsilon}} + \dotsb + r_{M-N_{\epsilon}}s_{N_{\epsilon}} \leq (M-2N_{\epsilon})(rs + \epsilon)
$$
(Remember $M\geq 2N_{\epsilon}$, so all the factors in the expression in the middle are close to their limits.) When we divide this by $M$, we find that as $M\to\infty$, $\dfrac{(2)}{M+1}$ is arbitrarily close to $rs$. Therefore, ( *) should converge to $rs$, since $\dfrac{(1)}{M+1}$ and $\dfrac{(3)}{M+1}$ are negligible.
