If $\sum a_n$ is a convergent series with $S = \lim s_n$, where $s_n$ is the nth partial sum, then $\lim_{n \to \infty} \frac{s_1+...+s_n}{n} = S$ Let $\sum a_n$ be a convergent series, and let $S = \lim s_n$, where $s_n$ is the nth partial sum. 
I need to prove the following:  
$\lim_{n \to \infty} \frac{s_1+...+s_n}{n} = S$  
How do I go about proving that proof?
Definition of a limit 
$\lim_{n \to \infty} f(x) = L$ if for every number $\epsilon>0$ there is some number $\sigma >0$ such that $|f(x)-L| <\epsilon$ whenever $0<|x-a|<\sigma$
 A: Let $\epsilon>0.$ Since $s_n\rightarrow S$ we can choose an $N$ such that $s_n>S-\epsilon$ for all $n>N.$ Let $n>N$. Then can write $$ \frac{s_1 + \ldots + s_n}{n} = \frac{s_1 + \ldots + s_N}{n} +\frac{s_{N+1}+\ldots s_n}{n} > \frac{s_1 + \ldots + s_N}{n}  + \frac{n-N}{n}(S-\epsilon).$$ Taking $\liminf$ of both sides gives $$\liminf_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \ge (S-\epsilon).$$ This is true for all $\epsilon>0,$ so we have $$\liminf_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \ge S.$$ 
By the same reasoning, can show that $$\limsup_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \le (S+\epsilon)$$ for any $\epsilon >0.$ so that $$\limsup_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \le S.$$ Since $\liminf \le \limsup,$ these inequalities yield $\liminf = \limsup = S,$ so the limit exists and it is equal to $S$.
A: Using the Stolz-Cesaro Theorem, we have
$$\lim_{n\to \infty}\frac{\sum_{k=1}^n\left(\sum_{j=1}^ka_j\right)}{n}=\lim_{n\to \infty}\sum_{j=1}^{n+1}a_j=S$$

If one is unfamiliar with Stolz-Cesaro, then on can proceed as follows.  Let $\epsilon>0$ be given.  Then, choose and fix $N$ so large that $|s_k-S|<\epsilon/2$ whenever $k>N$..  
Thus, we can write
$$\begin{align}
\left|\frac{\sum_{k=1}^n s_k}{n}-S\right|&=\left|\frac1n \sum_{k=1}^n (s_k-S)\right|\\\\
&\le \frac1n \sum_{k=1}^N |s_k-S|+\frac1n \sum_{k=N+1}^n |s_k-S|\\\\
&\le \frac1n \sum_{k=1}^N |s_k-S| +\left(1-\frac{N}{n}\right)\frac{\epsilon}{2}\\\\
&\le \frac1n \sum_{k=1}^N |s_k-S| +\frac{\epsilon}{2}\\\\
&\le \frac{\epsilon}{2}+\frac{\epsilon}{2}\\\\
&=\epsilon
\end{align}$$
whenever $n>\max\left(N+1,\frac{2\sum_{k=1}^N |s_k-S|}{\epsilon}\right)$
And we are done!
A: The point is that:
$$\big\lvert\frac{s_1+s_2+\ldots+s_n}{n}-S\big\rvert=\big\lvert\frac{s_1-S+s_2-S+\ldots+s_n-S}{n}\big\rvert\le\frac{1}{n}(\lvert s_1-S\rvert+\lvert s_2\rvert-S+\ldots+\lvert s_n-S\rvert)=\frac{1}{n}\sum_{i=1}^N\lvert s_i-S\rvert+\frac{1}{n}\sum_{i=N+1}^n\lvert s_i-S\rvert$$
The first term goes to zero when $n\rightarrow\infty$
By choosing $N$ large enough one can bound each of $\frac{1}{n}\sum_{i=1}^N\lvert s_i-S\rvert$ and $\frac{1}{n}\sum_{i=N+1}^n\lvert s_i-S\rvert$ by $\frac{\epsilon}{2}$
A: $$\left|\frac{1}{n}\sum_{k=1}^n s_k - \sum_{k=1}^n a_k\right| =\left|\sum_{k=1}^n (\frac{n-k+1}{n} a_k - a_k) \right|$$
$$ \leq \varepsilon \left|  \sum_{k=1}^m a_k\right| + \left| \sum_{k=m+1}^n a_k\right|\leq \varepsilon (|S|+\delta) + \varepsilon'$$
where for any given $\varepsilon,\varepsilon',\delta >0\ $, $m$ can be chosen so that the above holds for all sufficiently large $n$. Namely: First choose $m$ large enough so that  $\left|\sum_{k=1}^m a_k - S\right| < \delta$ and $\left| \sum_{k=m+1}^n a_k\right|<\varepsilon'$. Next, choose $n_0$ large enough so that for all $n\geq n_0$ it is true that  $ |\frac{1-k}{n}| < \varepsilon$ for $k\leq m$. Thus, the difference of  the two series becomes arbitrarily small; since one of them converges to $S$, the other one also converges to $S$.
A: You re-prove Cesaro's theorem. By definition there exists $N\in \mathbf N$ such that 
$$\lvert S_n-S\rvert <\varepsilon\quad\forall n>N. $$
We deduce that, for $n>N$, 
\begin{align}
\biggl\lvert\frac{S_1+\dots+S_n}n-S\,\biggr\rvert&=\biggl\lvert\frac{S_1-S+\dots+S_N-S}n+\frac{S_{N+1}-S+S_n-S}n\,\biggr\rvert\\
&\le\biggl\lvert\frac{S_1-S+\dots+S_N-S}n\biggr\rvert+\biggl\lvert \frac{S_{N+1}-S}n\biggr\rvert+\dots+\biggl\lvert \frac{S_n-S}n\,\biggr\rvert\\
&\le\biggl\lvert\frac{S_1+\dots+S_N}n-\frac{NS}n\biggr\rvert+\biggl\lvert \frac{S_{N+1}-S}n\biggr\rvert+\dots+\biggl\lvert \frac{S_n-S}n\,\biggr\rvert\\
&\le\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{\lvert S_{N+1}-S\rvert}n+\dots+\frac{\lvert S_n-S\rvert}n\\
&\le\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{\varepsilon}n+\dots+\frac{\varepsilon}n
=\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{(N-n)\varepsilon}n\\
&\le \frac{\lvert S_1+\dots+S_N+NS\rvert}n+\varepsilon\le 2\varepsilon
\end{align}
if $n$ is large enough.
