How to prove that if $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$ I'm wondering if I have a sufficient proof of the following: 
If $(a_n)$ is a sequence such that $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$. 
My approach:
For all $\varepsilon > 0$, there exists $N$ such that for all $k>N$, $|a_k -A|<\varepsilon$. So we can break the limit up as follows
$$\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A$$
Is this on the right track, or am I missing something about breaking up the limit in the way I have?

Since $(a_n)$ is convergent hence it's bounded. Let $|a_n|\leq K\ \forall\ n\in \mathbb{N}$. Now for $\epsilon >0$ let $N\in \mathbb{N}$ be such that $|a_n-A|<\epsilon\ ,\forall\ n>N$.
Consider $|\frac{a_1+a_2+...+a_n}n-A|=|\frac{(a_1-A)+(a_2-A)+...+(a_n-A)}n|\leq\frac{|a_1-A|}n+\frac{|a_2-A|}n+...+\frac{|a_n-A|}n$.
Now choose $M\in \mathbb{N}$ (What $M$ ?) and bound the first $M$ terms of the above expression using the boundedness of $(a_n)$ and the rest of the terms using the fact that $|a_n-A|\rightarrow0$ as $n\rightarrow \infty$. 
