Find the limit $\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}$ When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit,
$$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$
Thank you for your help in advance.

By the Cesàro mean theorem, if $(x_n)_{n\in\mathbb{N}^*}\to a$ then $\left(\bar{x}_n=\frac{1}{n}\sum_{j=1}^{n}x_j\right)_{n\in\mathbb{N}^*}\to a$.
So, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that all the quantities: 
$$|x_m-a|,\quad|y_m-b|,\quad|\bar{x}_m-a|,\quad|\bar{y}_m-b|$$ 
are less than $\epsilon$ for any $m\geq N$. If we set:
$$ c_n = \frac{1}{n}\sum_{i=1}^{n}x_i y_{n+1-i}, $$
for any $n\geq N$ we have that:
$$ c_{2n} = \frac{1}{2n}\sum_{j=1}^{n} x_j y_{2n+1-j}+\frac{1}{2n}\sum_{j=1}^{n} y_j x_{2n+1-j} $$
differs from $\frac{1}{2}b \bar{x}_n+\frac{1}{2}a \bar{y}_n$ no more than $\frac{\epsilon}{2}(|\bar{x}_n|+|\bar{y}_n|)$, so:
$$ \left(c_{2n}\right)_{n\in\mathbb{N}^*}\to ab. \tag{1}$$
In a similar fashion, for any $n\geq N$
$$ c_{2n+1} = \frac{2n}{2n+1}\left(\frac{1}{2n}\sum_{j=1}^{n} x_j y_{2n+2-j}+\frac{1}{2n}\sum_{j=1}^{n} y_j x_{2n+2-j}\right)+\frac{x_{n+1}y_{n+1}}{2n+1} $$
cannot differ from $\frac{2n}{2n+1}\left(\frac{1}{2}b \bar{x}_n+\frac{1}{2}a \bar{y}_n\right)$ more than $\left(\frac{\epsilon}{2}+\frac{\varepsilon^2}{n}\right)\cdot(|\bar{x}_n|+|\bar{y}_n|)$, so:
$$ \left(c_{2n+1}\right)_{n\in\mathbb{N}}\to ab. \tag{2}$$
Now $(1)$ and $(2)$ simply give:
$$ \left(c_n\right)_{n\in\mathbb{N}^*}\to ab \tag{3}$$
as expected.
