arithmetic mean of a sequence converges

There are still some typos in your proof (and it isn't written down clearly, in my opinion).

So your proof seems to use this idea: Let $\varepsilon>0$, then there exists $M \in \mathbb{N}$ such that for all $n \geq M: |a_n-a| \leq \frac{\varepsilon}{2}$. Thus

$$|\bar{a}_n-a| \leq \frac{1}{n} \underbrace{\sum_{k=1}^{M-1} |a_k-a|}_{=:N} + \frac{1}{n} \underbrace{\sum_{k=M}^n |a_k-a|}_{\leq \frac{1}{2} \varepsilon \cdot n}$$

Now let $n \geq \max \{M, \frac{2}{\varepsilon} \cdot N\}=:M'$, then $|\bar{a}_n-a|\leq \varepsilon$.

Whether you take the first sum from $k=1$ to $M$ or to $M-1$ doesn't matter. Important is that

• the first sum is finite and does not depend on $n$
• in the second sum there are only $|a_k-a|$ such that $k$ is bigger or equal than $M$.

You could also sum from $k=1$ to $M+j$ for some $j \geq 0$.

Moreover, choose $K := \frac{2}{\varepsilon} \cdot N$, then

$$\frac{1}{n} \underbrace{\sum_{k=1}^{M-1}|a_k-a|}_{N} \leq \frac{1}{K} \cdot N \leq \frac{\varepsilon}{2}$$

for all $n \geq K$, so the answer to your (first) question is "yes".

Since $\{a_n\}$ converges, it is bounded, and so is $\{a_n-a\}$. Choose $M$ such that $$ |a_n-a|\le M\quad\forall n\in\mathbb{N}. $$ If $1\le m<n$, then $$ \Bigl|\frac1n\sum_{k=1}^na_k-a\Bigr|\le\frac1n\sum_{k=1}^m|a_k-a|+\frac1n\sum_{k=m+1}^n|a_k-a|\le M\,\frac{m}{n}+\frac1n\sum_{k=m+1}^n|a_k-a|. $$ Let $\epsilon>0$ be given. First choose $m$ such that $|a_n-a|\le\epsilon/2$ for all $n>m$. Once $m$ is chosen, choose $n_0>m$ such that $M\,m/n_0\le\epsilon/2$. Observe that $n_0$ depends $m$ that depends on $\epsilon$, so hat $n_0$ depends on $\epsilon$. Then, if $n\ge n_0$ we have $$ \Bigl|\frac1n\sum_{k=1}^na_k-a\Bigr|\le M\,\frac{m}{n}+\frac1n\sum_{k=m+1}^n|a_k-a|\le\frac\epsilon2+\frac{n-m}{n}\frac\epsilon2\le\epsilon. $$