how to prove if $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$? I know this is a duplicate but I can't understand any of the duplicates.

$\lim_{n \rightarrow \infty} a_n = A$, how to prove if $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$?

If $a_n=A$ for all $n$, it would be easy, right? Now what if $A-\epsilon < a_n < A+\epsilon$ for all $n$? You could get that $|\frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$. If the limit exists, it is similarly bounded. To avoid saying "if the limit exists", take the limsup, which always exists, and say $|\limsup \frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$.
So here's the problem. You only have $|a_n-A|< \epsilon$ for $n \geq N$. What to do? Split it into two cases and work the rest from there:
$$\frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^N a_k + \frac{1}{n}\sum_{k=N+1}^n a_k.$$
I know that for large enough k the terms go to zero but what about the terms that are 1 to N? I can't figure out why.
 A: Put 
$$A_n=\frac{1}{n}\sum_{k=1}^na_k$$
Given $\varepsilon>0$, pick $N$ such that $|a_k-A|<\varepsilon/2$ for $k\ge N$. Since this $N$ is fixed, we can find $N'$ such that $$\frac{1}{N'}<\frac{\varepsilon}{2\left(\sum_{k=1}^N|a_k-A|+1\right)}$$
Now write, for $n\ge N+1$:
$$A_n-A=\frac{1}{n}\sum_{k=1}^n(a_k-a)=\frac{1}{n}\sum_{k=1}^N(a_k-a)+\frac{1}{n}\sum_{k=N+1}^n(a_k-a)$$
Then, if $n\ge\max\{N+1,N'\}$:
$$|A_n-A|\le \frac{1}{n}\sum_{k=1}^N|a_k-a|+\frac{1}{n}\sum_{k=N+1}^n|a_k-a|$$
$$\le \frac{1}{N'}\sum_{k=1}^N|a_k-a|+\frac{1}{n}(n-(N+1)+1)\frac{\varepsilon}{2}$$
$$\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
A: Let $\varepsilon>0$. Since $a_n\to A$, there exists $N\in \Bbb N$ such that $\forall n\geq N$, we have $|a_n-A|<\varepsilon$. $(*)$
The idea of the proof is to deal separately with the two sets of points: $a_n$ when $n<N$ and $n\geq N$.
\begin{align*}\left|\frac{a_1+\cdots+a_n}{n}-A\right|&\leq \frac{1}{n}(|a_1-A|+\cdots+|a_n-A|)\\
&=\frac{1}{n}\bigg(\underbrace{(|a_1-A|+\cdots+|a_{N-1}-A|)}_{(1)}+\underbrace{(|a_N-A|+...+|a_n-A|)}_{(2)}\bigg)
\end{align*}
$(1)$ can be dealt with using the fact that a convergent sequence is bounded and also that $\lim\limits_{n\to \infty} \frac{1}{n}=0$ (proof by Archimedean property).
$(2)$ can be made as small as we wish by $(*)$.
As a further exercise, one can try to show that given $a_n\to A$ and $b_n\to B$, then $\lim\limits_{n\to \infty} \dfrac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n}=AB$.
