two converging sequences results in third can anyone help me with this, I got stuck on it.
let $a_n$ and $b_n$ (for $n$ a non-negative integer) be two sequences of real numbers such that $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$. prove that $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.

Write $\Delta_n=a_n-a$ and $\delta_n=b_n-b$; clearly these each converge to $0$. Then
$$\frac{1}{n}\left(\sum_{k=0}^na_kb_{n-k}\right)-ab=\frac{1}{n}\left(\sum_{k=0}^n \left((\Delta_k+a)b_{n-k}-ab\right)\right)=\frac{1}{n}\left(\sum_{k=0}^n\left(\Delta_k b_{n-k}+a\delta_{n-k}\right)\right).$$
Since $b_i$ converges we can say it is bounded in magnitude by $B$. All you have to prove now is that
$$\lim_{m\to\infty}c_m=0\implies\frac{c_0+c_1+\cdots+c_m}{m}\to0. $$
