Problem
Given $\lim\limits_{n \to \infty}a_n=a$ and $\lim\limits_{n \to \infty}b_n=b$, prove that $$\lim_{n \to \infty}\frac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n}=ab.$$
My Proof
Denote $x_n=a_n-a,~~~y_n=b_n-b.$ Then $$\lim\limits_{n \to \infty}x_n=\lim\limits_{n \to \infty}y_n=0,\tag1$$ $$\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_k}{n}=\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n y_k}{n}=0,\tag2$$ $$\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_k^2}{n}=\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n y_k^2}{n}=0.\tag3$$ Moreover, since
$$0\leq\left|\vcenter{\frac{\sum\limits_{k=1}^n x_ky_{n-k+1}}{n}}\right|\leq\vcenter{\frac{\sum\limits_{k=1}^n |x_ky_{n-k+1}|}{n}}\leq\vcenter{\frac{\sum\limits_{k=1}^n (x_k^2+y_{k}^2)}{2n}}\to 0~~~(n \to \infty),$$ then $$\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_ky_{n-k+1}}{n}=0.\tag4$$ Therefore, \begin{align*} \lim_{n \to \infty}\frac{\sum\limits_{k=1}^n a_kb_{n-k+1}}{n}&=\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n [(x_k+a)(y_{n-k+1}+b)]}{n}\\ &=\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_ky_{n-k+1}+b\sum\limits_{k=1}^n x_k+a\sum\limits_{k=1}^n y_{n-k+1}+\sum\limits_{k=1}^n ab}{n}\\ &=\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_ky_{n-k+1}}{n}+b\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n x_k}{n}+a\lim_{n \to \infty}\frac{\sum\limits_{k=1}^n y_{n-k+1}}{n}+\lim_{n \to \infty}ab\\ &=0+b\cdot 0+a\cdot 0+ab\\ &=ab. \end{align*}