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?
 A: 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 $n\rightarrow \infty$. 
A: For any given $\epsilon>0$, choose $N$ such that $A-\epsilon/2<a_k<A+\epsilon/2$ for all values of $k>N$.  Then, whenever $n>\max\left(N,\frac{2\left|\sum_{k=1}^N(a_k-A)\right|}{\epsilon}\right)$ we have
We have
$$\begin{align}
\left|\frac1n\sum_{k=1}^na_k-A\right|&=\left|\frac1n\sum_{k=1}^n(a_k-A)\right|\\\\
&\le \left|\frac1n\sum_{k=1}^N(a_k-A)\right|+\left|\frac1n\sum_{k=N+1}^n(a_k-A)\right|\\\\
&\le \left|\frac1n\sum_{k=1}^N(a_k-A)\right|+\frac1n\sum_{k=N+1}^n\left|a_k-A\right|\\\\
&\le \frac{\epsilon}{2}+\frac{\epsilon}{2}\left(1-\frac Nn\right)\\\\
&< \epsilon
\end{align}$$
And we are done!
A: Given $e>0,$ choose the least (or any) $k$  such that $i>k \implies |a_i-A|<\frac {e}{3}.$ 
Now choose $m$   large enough that $m>k$ and  $\frac {k}{m}|A|<\frac {e}{3}$ and $m^{-1}|\sum_{i=1}^ka_i|<\frac {e}{3} .$ 
For $n\geq m$  we have $$|-nA+\sum_{i=1}^na_i|= |\sum_{i=1}^ka_i+kA+\sum_{i=k+1}^n(a_i-A)|<$$ $$<\frac {em}{3}+k|A|+\sum_{i=k+1}^n|a_i-A|<\frac {em}{3}+k|A|+(n-k)\frac {e}{3}.$$  So $n\geq m$ implies $$|-A+n^{-1}\sum_{i=1}^na_i|\; <\; 
\frac {e}{3} \frac {m}{n}+\frac {m}{n}\cdot \frac {k}{m}|A|+(1-k/n)\frac {e}{3}<e.$$
