A sequence and convergence Let $\{x_n\}$ and $\{y_n\}$ be sequences of real numbers which converge to $\ell$ and $m$ respectively. Show that 
$$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}=\ell m$$
This is a problem I found in a book, it seemed to be routine, but now I can't solve it. 
What I tried :

Let us take some $\epsilon>0$. Now after a stage $N$, both $|x_n-\ell|<\epsilon$ and $|y_n-m|<\epsilon$. Now I wanted to break the sum $$\left|\frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}-\ell m\right|$$in two parts, one upto $N$, and other one for $N+1$ to $n$. Now we need this to be arbitrarily small. But here I am lost. I don't know how to proceed anymore. Because when I take $x_i's$ to make up a finite sum, then the $y_j's$ are there too with $j$ very large. I think this is the main idea, but unable to do anything further. Can someone help me? Thanks.

 A: First, recall that if $x_n\to x$, and if $\widehat{x_n}=\displaystyle \frac 1 n\sum_{i=1}^n x_i$ then $\widehat{x_n}\to x$. Now, you have $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}=xy$$
Subtract $\displaystyle y\frac 1n\sum_{k=1}^n x_k $ to get $$\frac 1 n\sum_{i=1}^nx_k(y_{n+1-k}-y)$$
It suffices we show this converges to $0$, since $\displaystyle y\frac 1n\sum_{k=1}^n x_k \to xy$, that is, we may assume that $y_i\to 0$. By subtracting $\displaystyle x\frac 1n \sum_{k=1}^n(y_{n+1-k}-y)$ we may as well assume $x_i\to 0$.
Given $\varepsilon >0$, pick $M>0$ and $n>M$, and write $$\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}{y_{n + 1 - i}}}  = \frac{1}{n}\sum\limits_{i = 1}^M {{x_i}{y_{n + 1 - i}}}  + \frac{1}{n}\sum\limits_{i = M + 1}^n {{x_i}{y_{n + 1 - i}}} $$
Now choose $M$ such that $i>M$ gives $|x_i|<\varepsilon/B$ where $B$ is a bound for $|y_i|$ and such that $i>M$ gives $|y_i|<\varepsilon/A$ where $A$ is a bound for $|x_i|$. Then when $n>2M$, $$\eqalign{
  & \frac{1}{n}\sum\limits_{i = 1}^M {\left| {{x_i}} \right|\left| {{y_{n + 1 - i}}} \right|}  + \frac{1}{n}\sum\limits_{i = M + 1}^n {\left| {{x_i}} \right|\left| {{y_{n + 1 - i}}} \right|}  \leqslant   \cr 
  & \frac{A}{n}\sum\limits_{i = 1}^M {\left| {{y_{n + 1 - i}}} \right|}  + \frac{B}{n}\sum\limits_{i = M + 1}^n {\left| {{x_i}} \right|}  \leqslant   \cr 
  & \frac{A}{n}\frac{\varepsilon }{A} + \frac{B}{n}\left( {n - M} \right)\frac{\varepsilon }{B} = M\frac{\varepsilon }{n} + \left( {n - M} \right)\frac{\varepsilon }{n} = \varepsilon  \cr} $$
A: first reduce to $x_k, y_k\to 0$ (for simplicity).  let $|x_k|\leq M, |y_k|\leq N$.  then 
$$
\frac{1}{2n}\sum_{k=0}^{2n}x_ky_{2n-k}\leq M\max_{k\geq n}\{\delta_k\}+N\max_{k\geq n}\{\epsilon_k\},
$$
which goes to zero as $n\to\infty$.
(i'm just splitting the list $\epsilon_k\delta_{n-k}$ in half, which bounds the index of one of the factors below by something that goes to infinity as $n$ does.)
A: As you have said, we consider two cases:
Case 1: if $l=0$, for $\epsilon >0, \exists N\in\mathbb{N}, st |x_n|<\epsilon,\forall n\geq N .$ Then $\bigg |\frac{1}{n}\sum_{k=1}^n x_ky_{n+1-k}-0\bigg|=\bigg |\frac{1}{n}(\sum_{k=1}^N x_ky_{n+1-k}+\sum_{k=N+1}^n x_ky_{n+1-k})\bigg|<\frac{1}{n}(\sum_{k=1}^N |x_k|\cdot \sum_{k=1}^N y_{n+1-k})+\frac{\epsilon}{n} \sum_{k=1}^N |y_k|\rightarrow 0, n\rightarrow \infty$
(Notice that $y_k\rightarrow m$, so $\sum_{k=1}^N y_{n+1-k}$ is bounded)
Case 2: if $l\neq 0$, $\bigg|\frac{1}{n}\sum_{k=1}^n x_ky_{n+1-k}-lm\bigg|=\bigg|\frac{1}{n}\sum_{k=1}^n (x_k-l)y_{n+1-k}+\frac{1}{n}\sum_{k=1}^n l\cdot y_{n+1-k}-lm\bigg|=\bigg|\frac{1}{n}\sum_{k=1}^n (x_k-l)y_{n+1-k}+l\cdot(\frac{1}{n}\sum_{k=1}^n  y_{n+1-k}-m)\bigg| \leq \bigg|\frac{1}{n}\sum_{k=1}^n (x_k-l)y_{n+1-k} \bigg|+l\cdot \bigg| \frac{1}{n}\sum_{k=1}^n y_k-m\bigg|\rightarrow 0, n\rightarrow \infty$.
($ \bigg|\frac{1}{n}\sum_{k=1}^n (x_k-l)y_{n+1-k} \bigg|\rightarrow 0, n\rightarrow \infty $because of case 1)
