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: You could just apply Stolz-Cesaro blindly or to gain some insight make an argument along the following lines.
Since $a_n \to l$, for any $\epsilon>0$ there is a positive integer $N$ such that $l- \epsilon < a_n < l + \epsilon$ when $n > N$.
Hence, with $S_n = a_1 + a_2 + \ldots + a_n$,
$$\frac{1}{n}S_n = \frac1{n}\sum_{k=1}^{N}a_k + \frac1{n}\sum_{k=N+1}^{n }a_k \\\leqslant \frac1{n}\sum_{k=1}^{N}a_k + \frac1{n}(n-N)(l + \epsilon),$$
and
$$\limsup_{n \to \infty}\frac{1}{n}S_n \leqslant l + \epsilon.$$
Make a similar argument to show that 
$$\liminf_{n \to \infty}\frac{1}{n}S_n \geqslant l - \epsilon.$$
Since both inequalities hold for any $\epsilon> 0$ we have
$$\limsup_{n \to \infty}\frac{1}{n}S_n = \liminf_{n \to \infty}\frac{1}{n}S_n = \lim_{n \to \infty}\frac{1}{n}S_n=l$$
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 $\epsilon/2$.
A: Let $x_n=a_1+a_2+...+a_n$ and $y_n=n$. Now, $\displaystyle \lim_{n\rightarrow \infty}\frac {x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\rightarrow \infty}a_{n+1}=\lim_{n\rightarrow \infty}a_{n}=l$. So by Cesaro-Stolz’s theorem, $\displaystyle \lim_{n\rightarrow \infty}\frac{x_n}{y_n}=\lim_{n\rightarrow \infty}\frac{a_1+a_2+...+a_n}{n}=l.$
A: Note that
$$\lim_{n\to 0}a_n=\lim_{n\to 0}\frac{\sum_{k=1}^na_k-\sum_{k=1}^{n-1}a_k}{(n)-(n-1)}=\ell$$
By the Stolz-Cesaro Theorem we have immediately that
$$\lim_{n\to 0}\frac{\sum_{k=1}^na_k}{n}=\ell$$
