Let $\lim_{n\to\infty}a_n$ = a, $\lim_{n\to\infty}b_n$ = b. We need to find $\lim_{n\to\infty}\frac{a_nb_1+a_{n-1}b_2+...+a_1b_1}{n}$. Let $\lim_{n\to\infty}a_n$ = a, $\lim_{n\to\infty}b_n$ = b. We need to find $\lim_{n\to\infty}\frac{a_nb_1+a_{n-1}b_2+...+a_1b_1}{n}$. 
As far as I get it, the answer is $ab$. 
I've tried this to prove it: as both $a_n$ and $b_n$ have a limit, they also both have a supremum. Let $C_1$ and $C_2$ be the supremua. Then by the properties of supremum for any given $е$ there are $k, l$ such that 
$$\frac{a_nb_1+a_{n-1}b_2+...+a_1b_n}{n}<C_1C_2<(a_k+e)(b_l+e)$$
And I feel like there is some tiny step from here to the solution as the above is more or less $ab$, but I just can't get there. Would you mind giving some hints?

$$\lim_{2n\to\infty} \frac{a_{2n}b_1+\cdots+a_{n+1}b_n+a_nb_{n+1}+\cdots+a_1b_{2n}}{2n}=$$
$$\lim_{2n\to\infty} \frac{a_{2n}b_1+\cdots+a_{n+1}b_n}{2n} + \frac{a_nb_{n+1}+\cdots+a_1b_{2n}}{2n}=$$
$$\lim_{2n\to\infty} (a+\epsilon(2n))\frac{b_1+\cdots+b_n}{2n}+(b+\epsilon'(2n))\frac{a_1+\cdots+a_n}{2n}$$
Where $\epsilon, \epsilon'$ each go to zero with $n$.
$$=a\frac{b}{2}+b\frac{a}{2}=ab$$
Using Cesaro mean as Andrew suggests.
