If $a_n\to \ell $ then $\hat a_n\to \ell$ [duplicate]

For the tail sum $$\frac1n\sum_{N+1}^n|a_k-\ell|\leq \frac{(n-N)}{n}\varepsilon<\varepsilon$$ for all large $n$, hence $$\limsup_n \frac1n\sum_{1}^n|a_k-\ell|\leq 0+\varepsilon=\varepsilon...$$

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 $$\hat a_n\to\ell.$$