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.

In the special case that $x_n,y_n\ge0$, we can prove the statement as follows: by the Arithmetic Mean-Geometric Mean (first inequality)  and the Cauchy-Schwarz ( second inequality)  we can bound the given term as follows:
$$\sqrt[n]{x_1\ldots x_ny_1\ldots y_n}\le \dfrac{x_1y_1+\ldots+x_ny_n}{n} \le \dfrac{\sqrt{x_1^2+\ldots+x_n^2}\sqrt{y_1^2+\ldots+y_n^2}}{n}$$
Letting $n \to \infty$ we have that : $$\sqrt[n]{a^nb^n}\le \lim_{n \to \infty} \dfrac{x_1y_1+\ldots+x_ny_n}{n}\le\sqrt{a^2}\sqrt{b^2}\tag{1}$$ or equivalently $$ab\le \lim_{n \to \infty} \dfrac{x_1y_1+\ldots+x_ny_n}{n}\le ab$$
Using (1),  if $x_n \to a$, as $n \to \infty,$
 then also we should have
$$\prod_{k=1}^{n} x_k^{1/n}\to a \quad \text{as} \quad n\to \infty, $$ 
since $$\prod_{k=1}^{n}x_k^{1/n}=\exp\left(\frac{1}{n}\sum_{k=1}^{n}\ln x_k\right)$$ and $\exp, \ln$ are continuous, thus enabling taking the limit taken inside. We also used the well known fact that if $x_n \to a$ as $n \to \infty$ then $x_n^2  \to a^2$ and also$$\frac{x^2_1+\ldots+x^2_n}{n} \to a^2$$ as $n \to \infty$.
