If $a_n\to \ell $ then $\hat a_n\to \ell$ 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:
$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}}  = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}}  - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = \ell $$
 A: For the tail sum 
$$\frac1n\sum_{N+1}^n|a_k-\ell|\leq \frac{(n-N)}{n}\varepsilon<\varepsilon$$
for all large $n$, hence
$$\limsup_n \frac1n\sum_{1}^n|a_k-\ell|\leq 0+\varepsilon=\varepsilon...$$
A: Just rewriting your argument:
The sequence $\{|a_n-\ell|\}$ converges to $0$ therefore it's bounded, so there exist some $M\gt 0$ such that
$$|a_n-\ell|\leq M\quad\forall n\in \Bbb N.$$
This $M$ plays the role  of the $\zeta$ in your proof, but notice it does not depends on $n$.
Let $\epsilon\gt 0$. There exist $n_1\in\Bbb N$ such that for all $n\in\Bbb N$, $n\geq n_1$ implies $$|a_n-\ell|\lt\frac{\epsilon}{2}.$$
Since $$\lim_{n\to\infty} \frac{n_1}{n}=0,$$ there exist $n_2\in\Bbb N$ such that, for all $n\in\Bbb N$, $n\geq n_2$ implies $$\frac{n_1}{n}\leq \frac{\epsilon}{2M}.$$
Let $N=\max\{n_1,n_2\}$, then for all $n\in\Bbb N$, $n\geq N\geq n_1$ implies
$$\left(1-\frac{n_1}{n}\right)\leq 1$$
and then
$$\begin{align*}
\left| -\ell+\frac1{n}\sum_{j=1}^n a_j \right| &\leq \frac1{n}\sum_{j=1}^n \left|a_j-\ell\right|\\
&= \frac1{n}\sum_{j=1}^{n_1} \left|a_j-\ell\right| + \frac1{n}\sum_{j=n_1+1}^{n} \left|a_j-\ell\right|\\
&\lt \frac1{n} n_1M + \frac1{n}(n-n_1)\frac{\epsilon}{2}\\
&=\frac{n_1}{n}M + \left(1-\frac{n_1}{n}\right)\frac{\epsilon}{2}\\
&\lt \frac{\epsilon}{2M}M + 1\cdot \frac{\epsilon}{2}=\epsilon.
\end{align*}$$
Therefore $$\hat a_n\to\ell.$$
