I need some help to finish this proof:
THEOREM
Let $\{a_n\}$ be such that $\lim a_n=\ell$ and set
$\hat a_n=\frac 1 n \sum_{k=1}^na_k$
Then $\lim\hat a_n=\ell$
PROOF
Let $\epsilon >0 $ be given. Since $\lim a_n=\ell$ , there exists an $N$ for which $\left| {{a_n} - \ell } \right| < {\epsilon/2 }$
whenever $n>N$. Now:
$\begin{eqnarray*} \left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^N {\left( {{a_k} - \ell } \right)} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &\leqslant& \frac{1}{n}\sum\limits_{k = 1}^N {\left| {{a_k} - \ell } \right|} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} \\ & <& \frac{N}{n}\zeta - \frac{{N }}{2n}\epsilon+\epsilon/2 \end{eqnarray*} $
where $\zeta=\mathop {\max }\limits_{1 \leqslant k \leqslant N} \left| {{a_k} - \ell } \right|$
Now, let $n_0$ be such that if $n>n_0$,
$\eqalign{ & \frac{{N\zeta }}{n} < {\epsilon} \cr & \frac{N}{n} < {1} \cr} $ Then we get
$\frac{N}{n}\zeta - \frac{N}{n}\frac{\epsilon }{2} + \frac{\epsilon }{2} < \epsilon - \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon $
How far is this OK? Do you think there is an easier way to go about proving it? I now remember that by Stolz Cesàro:
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = \ell $