Q:

If $\lim_{n\to\infty}a_{n}=l$, Then prove that $\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l$

Given $a_n$ be a sequence and IF $\lim_{n\to\infty}a_{n}=l$, Then prove that  $\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l$
I do not know how to do this. Can someone help me with this?
Thanks
ATTEMPT

A:

Hint: 
Given $\epsilon > 0$, find $N$ so that $|a_k-L|<\epsilon/2$ for $k > N$. Then
$$\left | \frac{\sum_{k=1}^n a_k}{n} - L \right | = \left | \frac{\sum_{k=1}^n a_k - L}{n} \right | \leq \frac{\sum_{k=1}^n |a_k-L|}{n}.$$
By how this was already set up, the contribution to the average from the terms with $k > N$ is already less than $\epsilon/2$. Now try to take $n$ a bit larger, so that the contribution to the average from the terms with $k \leq N$ is also less than $\epsilon/2$.

