Showing that $ \sum \limits_{m=1}^{n} b_m x_{m-n}~\to~ ab$ as $n~\to~\infty$ If $x_n ~\to ~a$ as $n~ \to~ \infty$
Does:
$ \sum \limits_{m=1}^{n} b_m x_{n-m}~\to~ ab$ as $n~\to~\infty$?
$b_m ~\geq~0$ and $ b~\equiv~ \sum \limits_{m=1}^{\infty} b_m < \infty$
My attempt:
$\lim\limits_{n\to\infty}x_n~=~a$
I would like to have something like this I think
$$\inf x_n \sum_{m=1}^n b_m~\leq~\sum b_m x_{n-m}~\leq ~\sup x_n \sum_{m=1}^n b_m$$
Then taking the limits will squeeze it in to $ab$, I am having trouble getting this form, is there any easier way to do this?
i.e. $$\sum_{m=1}^n b_m\inf x_{n-m} ~\leq~\sum b_m x_{n-m}~\leq ~\sum_{m=1}^n b_m\sup x_{n-m} $$
Then $\sup x_{n-m} \leq \sup x_n$ and $\inf x_{n-m} \geq \inf x_n$
Is $b_m \geq 0$ necessary?
 A: Write
$$
\begin{align*}
\left| \sum_{m=1}^n b_m x_{n-m+1} - ab \right|
&= \left| \sum_{m=1}^n b_m x_{n-m+1} - a\sum_{m=1}^n b_m + a\sum_{m=1}^n b_m - ab \right|\\
&\leq \left|\sum_{m=1}^n b_m x_{n-m+1} - a\sum_{m=1}^n b_m \right| + \left|a\sum_{m=1}^n b_m - ab \right| \\
&\leq \left|\sum_{m=1}^n b_m \left( x_{n-m+1} - a\right) \right| + \left|a\left( \sum_{m=1}^n b_m - b\right)\right| \\
&\leq \left|\sum_{m=1}^n b_{n-m+1} \left( x_m - a\right) \right| + |a|\left| \sum_{m=1}^n b_m - b\right|          \tag{$\dagger$}
\end{align*}
$$
The second term goes to $0$ by convergence of the series $\sum b_m$, so it only remains to deal with the first one. Define the sequence $(a_n)_n$ by $a_n = x_n -a$ (so that it converges to $0$), and fix any $\varepsilon > 0$: there exists $N_\varepsilon\in\mathbb{N}$ such that $\forall n \geq N_\varepsilon$, $|a_n|\leq \varepsilon$.
Now, for any $n \geq N_\varepsilon$,
$$
\begin{align*}
\Delta_n\stackrel{\rm{}def}{=}\left|\sum_{m=1}^n b_{n-m+1} \left( x_m - a\right) \right| 
&= \left|\sum_{m=1}^n b_{n-m+1} a_m \right|
= \left|\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1} a_m + \sum_{N_\varepsilon}^{n} b_{n-m+1} a_m \right| \\
&\leq \left|\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1} a_m\right| + \left|\sum_{N_\varepsilon}^{n} b_{n-m+1} a_m \right| \\
&\leq \sum_{m=1}^{N_\varepsilon-1} \left|b_{n-m+1} a_m\right| + \sum_{N_\varepsilon}^{n} \left|b_{n-m+1} a_m \right| \\
&\leq \underbrace{\max_{1\leq k \leq N_\varepsilon-1}|a_k|}_{\alpha_{\varepsilon}}\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}  + \varepsilon\sum_{N_\varepsilon}^{n} b_{n-m+1} \qquad ((b_k)\text{ non-negative})\\
&= \alpha_{\varepsilon}\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}  + \varepsilon\sum_{N_\varepsilon}^{n} b_{n-m+1}
\end{align*}
$$
Finally, $0\leq \sum_{N_\varepsilon}^{n} b_{n-m+1} \leq \sum_{1}^{\infty} b_{n} = b$ (it actually converges when $n\to+\infty$ (monotone convergence) to some $b_\varepsilon\leq b$); while $\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}$ is the sum of constantly many ($N_\varepsilon-1$ to be precise) terms, each of them going to $0$ as $n\to+\infty$ (since $\sum b_k$ converges, $b_k\to 0$), and therefore $\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}\xrightarrow[n\to\infty]{}0$. Therefore, there exists $N^\prime_\varepsilon \geq N_\varepsilon$ such that for all $n\geq N^\prime_\varepsilon$, $\alpha_\varepsilon\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}\leq b\varepsilon$:
$$
\forall n\geq N^\prime_\varepsilon,\ \Delta_n\leq 2b\varepsilon
$$
i.e. $\Delta_n\xrightarrow[n\to\infty]{}0$. Thus the first term of ($\dagger$) goes to $0$ as well, proving that
$$
\left| \sum_{m=1}^n b_m x_{n-m+1} - ab \right|\xrightarrow[n\to\infty]{}0
$$
A: You can rewrite your sum as $$\sum_{m=0}^{n-1} b_{n-m} x_m$$ Define $$a_{n,m}=\frac{b_{n-m}}{\sum_{k=1}^n b_k}, m=0,\cdots,n$$ and $a_{n,m}=0$ for $m>n$. Then $\sum_{m=0}^\infty a_{n,m}=1$ for all $n$ and $\lim_{n\to\infty}a_{n,m}=0/b=0$. By Silverman-Toeplitz Theorem $$\lim_{n\to\infty}\sum_{m=0}^\infty a_{n,m}x_m=\lim_{n\to\infty}\frac{\sum_{m=0}^{n-1} b_{n-m} x_m}{\sum_{k=1}^n b_k}=a$$ Since $\lim_{n\to\infty}\sum_{k=1}^n b_k=b$, we get $$\lim_{n\to\infty}\sum_{m=0}^{n-1} b_{n-m} x_m=ab$$
