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 $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.

Solution 1:

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

$$\lim_{m\to\infty}c_m=0\implies\frac{c_0+c_1+\cdots+c_m}{m}\to0. $$

Solution 2:

We have, denoting $M:=\sup_{k\in\mathbb N}|a_k|<\infty$ \begin{align}\frac 1n\left|\sum_{k=0}^na_kb_{n-k}-ab\right|&\leq \frac 1n\left|\sum_{k=0}^na_k(b_{n-k}-b)\right|+\frac 1n\left|\sum_{k=0}^n(a_k-a)b\right|\\ &\leq \frac Mn\sum_{k=0}^n|b_k-b|+\frac{|b|}n\sum_{k=0}^n|a_k-a|, \end{align} hence for all $N\in\mathbb N$: $$\limsup_n\frac 1n\left|\sum_{k=0}^na_kb_{n-k}-ab\right|\leq \limsup_n\frac Mn\sum_{k=N}^n|b_k-b|+\limsup_n\frac{|b|}n\sum_{k=N}^n|a_k-a|,$$ and fixing $\delta>0$, and $N$ such that if $k\geq N$ then $\max\{|a_k-a|,|b_k-b|\}\leq \delta$ we get $$\limsup_n\frac 1n\left|\sum_{k=0}^na_kb_{n-k}-ab\right|\leq\limsup_n\frac{n-N+1}n(M+|b|)\delta=(M+|b|)\delta$$ and since $\delta$ is arbitrary we conclude that $\lim_{n\to\infty}\frac 1n\sum_{k=0}^na_kb_{n-k}=ab$.