$\lim x_n = a \in \mathbb{R}$, $\lim y_n = b \in \mathbb{R}$. Prove that $\lim \frac{x_1y_n + x_2y_{n-1} + \cdots + x_ny_1}{n} = ab$ So, here we have two converging sequences. It is obvious that from certain point our $x_n$ will be extremely close to $a$ and $y_n$ will be extremely close to $b$. Hence, we have infinite number of their reversed products which are extremely close to $ab$, there are only finite number of products for fixed $\epsilon$ which are not in $(ab - \epsilon, ab + \epsilon)$.
Okay, these are intuitive assumptions, but how to prove it formally?
 A: Suppose that
$$
\lim_{n\to\infty}a_n=a\tag1
$$
then, for any $\epsilon\gt0$, there is an $n_\epsilon$ so that $n\ge n_\epsilon\implies|a_n-a|\le\epsilon$. Therefore, for any $\epsilon\gt0$,
$$
\begin{align}
\limsup_{n\to\infty}\frac1n\sum_{k=1}^n|a_n-a|
&\le\limsup_{n\to\infty}\frac1n\sum_{k=1}^{n_\epsilon-1}|a_k-a|
+\limsup_{n\to\infty}\frac1n\sum_{k=n_\epsilon}^n|a_k-a|\tag2\\
&\le0+\epsilon\tag3
\end{align}
$$
Since $(3)$ is true for any $\epsilon\gt0$, we must have
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^n|a_n-a|=0\tag4
$$

We are given that $\lim\limits_{n\to\infty}a_n=a$ and $\lim\limits_{n\to\infty}b_n=b$. Since $b_n\to b$, there is some $B$ so that $|b_n|\le B$.
Therefore, since $a_{n-k+1}b_k-ab=(a_{n-k+1}-a)b_k+a(b_k-b)$, we have
$$
\begin{align}
\lim_{n\to\infty}\frac1n\sum_{k=1}^n|a_{n-k+1}b_k-ab|
&\le\lim_{n\to\infty}\frac1n\sum_{k=1}^n|a_{n-k+1}-a|\,|b_k|+\lim_{n\to\infty}\frac an\sum_{k=1}^n|b_k-b|\tag5\\
&\le\lim_{n\to\infty}\frac Bn\sum_{k=1}^n|a_k-a|+\lim_{n\to\infty}\frac an\sum_{k=1}^n|b_k-b|\tag6\\[6pt]
&=0\tag7
\end{align}
$$
Explanation:
$(5)$: triangle inequality
$(6)$: apply $|b_n|\le B$
$\phantom{\text{(6):}}$ substitute $k\mapsto n-k+1$ in the sum of $|a_{n-k+1}-a|$
$(7)$: apply $(4)$

Thus, $(7)$ leads us to
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^na_{n-k+1}b_k=ab\tag8
$$
A: It could be by This fact("theorem") that for a given array of sequences satisfying the regularity condition and a given convergent sequence $\{b_{n}\}$ with the limit $b$. Then the sequence $c_{n}$ defined as $\sum_{k=1}^{\infty} a_{n,k}b_{k}$ is convergent with the limit equal to $b$.
We consider two cases: First, one of the limits, let's say $a$, is zero. Second, None of the limits is zero.
For the first case, define $a_{n,k} = \frac{1+a_{n-k+1}}{n}$. (Where  $k \leq n$, other wise $0$.) It is easy to check that this array of sequences satisfy the regularity condition. Now the above theorem says that :
$$ \lim_{n \to \infty} \frac{(1+a_{1})(b_n)+...+(1+a_{n})(b_1)}{n} = b.$$
Since $\lim_{n \to \infty} \frac{b_1+...+b_n}{n} = b$, we can deduce the desired result.
Now let's deal with the other case.
Define $a_{n,k} = \frac{a_{n-k+1}}{na}$ and $0$ when ever $k >n$. Again this array satisfies the regularity condition. Thus the theorem says that :
$$ \lim_{n \to \infty} \frac{b_1a_{n}+...+b_{n}a_{1}}{na}=b.$$
And we get the desired result.
A: We have
\begin{align}
&\frac{x_1y_n + x_2y_{n-1} + \cdots + x_n y_1}{n}\\ 
=\ & \frac{(x_1 - a)y_n + (x_2 - a)y_{n-1} + \cdots + (x_n-a)y_1}{n} + a \frac{y_1 + y_2 + \cdots + y_n}{n}.\tag{1}
\end{align}
Since $\lim y_n = b$, we have $|y_n| \le C, \forall n$ where $C > 0$ is some absolute constant.
Then we have
\begin{align}
&\left|\frac{(x_1 - a)y_n + (x_2 - a)y_{n-1} + \cdots + (x_n-a)y_1}{n}\right|\\
\le\ & C \frac{|x_1 - a| + |x_2 - a| + \cdots + |x_n-a|}{n}.\tag{2}
\end{align}
Since $\lim |x_n - a| = 0$, by Stolz-Cesaro theorem (https://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf), we have
$$\lim \frac{|x_1 - a| + |x_2 - a| + \cdots + |x_n-a|}{n} = 0. \tag{3}$$
From (2) and (3), we have
$$\lim \frac{(x_1 - a)y_n + (x_2 - a)y_{n-1} + \cdots + (x_n-a)y_1}{n} = 0.$$
Also, since $\lim y_n = b$, by Stolz-Cesaro theorem, we have $\lim \frac{y_1 + y_2 + \cdots + y_n}{n} = b$.
Finally, we have $\lim \frac{x_1y_n + x_2y_{n-1} + \cdots + x_n y_1}{n} = ab$.
