Is it true that the $\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$? I came across a question which went like this:

If $a_n\to a$ and $b_n\to b$, then prove or disprove the following: $$\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$$

Here's how I approached the problem (I tried to prove it cause I thought that it is true):
We already have $a_n\rightarrow a$ and $b_n\rightarrow b$. Thus,
$$\begin{align}\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|&=\left|\frac{\sum_{i=1}^na_ib_{n-i+1}-nab}{n}\right|\\
&=\left|\frac{\sum_{i=0}^n\left(a_ib_{n-i+1}-ab\right)}{n}\right|\\
&=\frac{1}{n}\left|\sum_{i=1}^n\left(a_ib_{n-i+1}-ab\right)\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_ib_{n-i+1}-ab\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_i\right|\left|b_{n-i+1}-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|\end{align}$$
Now as $a_n$ is convergent$\implies\left|a_n\right|\leq M$ for some $M\in\mathbb{R}$. Thus,
$$\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|\leq\frac{M}{n}\sum_{i=1}^n\left|b_i-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|$$.
Now I am stuck as to how can I show that this whole expression is $<\epsilon$. Any hints as well as alternative methods are totally welcome.
NOTE: This might be a similar question only that it is a special case with $a=0$.
 A: Let $M> 0$ be such that $|a_i|, |b_i| \leq M$ for every $i$.
Given $\varepsilon > 0$, let $N\in\mathbb{N}$ be such that
$$
|a_i - a| < \varepsilon,\quad
|b_i - b| < \varepsilon, \qquad \forall i \geq N.
$$
Let $n > 2N$. We have that
$$
\sum_{i=1}^{n} a_i b_{n+1-i} = 
\left( \sum_{i=1}^N + \sum_{i=n-N+1}^n\right) a_i b_{n+1-i}
+ \sum_{i=N+1}^{n-N} a_i b_{n+1-i}\,.
$$
The first term at r.h.s. is bounded by $2 N M^2$.
Moreover, for $i = N+1,\ldots, n-N$ we have that
$$
|a_i b_{n+1-i} - ab| \leq M|a_i - a| + M |b_i - b| \leq 2 M \varepsilon,
$$
hence the second term is bounded by $2 n M \epsilon$.
All in all, we get that
$$
\left|\frac{1}{n} \sum_{i=1}^{n} (a_i b_{n+1-i} - ab)\right|
\leq \frac{4 N M^2}{n} + 2 M \varepsilon,
$$
so that
$$
\limsup_{n\to +\infty}
\left|\frac{1}{n} \sum_{i=1}^{n} (a_i b_{n+1-i} - ab)\right| \leq 2M \varepsilon
$$
and the required convergence follows.
A: I'm going to assume $a,b$ are finite.
Let $N>0$ such that $|b_n-b|<\epsilon$ and $|a_n-a|<\epsilon$ for any $n>N$. Then for $n>N$ we have $\sum_{i=1}^n|b_i-b|\leq B_N+\epsilon(n-N)$ and $\sum_{i=1}^n|a_i-a|\leq a_N +\epsilon(n-N)$ where here $A_N=\sum_{i=1}^N|a_i-a|$ while $B_N=\sum_{i=1}^N|b_i-b|$. This implies for $n>N$ that $$\Bigg|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\Bigg|\leq \frac{M}{n}\Bigg[B_N+\epsilon(n-N)\Bigg]+\frac{b}{n}\Bigg[A_N+\epsilon(n-N)\Bigg]$$ Take $n \longrightarrow \infty$ to see how $$\lim_{n\rightarrow \infty}\Bigg|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\Bigg|\leq\epsilon(M+b)$$
