Prove the limit of the sequence Assume that $\lim_{n \rightarrow \infty} a_{n} = A$
Prove that  $\lim_{n \rightarrow \infty} \frac{a_{1}+a_{2}+...+a_{n}}{n} = A$
I tried to seperate the sequence to 2 another sequences
$\lim_{n \rightarrow \infty} \frac{a_{1}+a_{2}+...+a_{k} + a_{k+1} + ... + a_{n}}{n} = A$
And then $\lim_{n \rightarrow \infty} \frac{Ck + (n-k)A}{n} = A$
But im stuck now, how are my stepes ?
There is some another way to proove that ?
Thanks.
 A: Let $\epsilon > 0$ be given. Choose $N_1$ large enough such that $|a_n - A| < \frac{\epsilon}{2}$ whenever $n > N_1$. Define
$$
M = \max\{|a_1 - A|, \ldots, |a_{N_1} - A|\},
$$
and choose $N_2$ large enough such that $\frac{1}{N_2} < \frac{\epsilon}{2 N_1 M }$.
Let $N = \max\{N_1, N_2\}$.
Then for $n > N$,
$$
\begin{aligned}
\left|\frac{a_1 + \cdots + a_n}{n} - A\right|
&\leq \frac{|a_1 - A| + \cdots + |a_n - A|}{n} \\
&= \frac{|a_1 - A| + \cdots + |a_{N_1} - A|}{n}
+ \frac{|a_{N_1+1} - A| + \cdots + |a_n - A|}{n} \\
& < \frac{N_1 M}{n}
+ \frac{(n - N_1) \epsilon}{n} \\
& < \frac{N_1 M}{N_2}
+ \frac{\epsilon}{2} \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
&= \epsilon.
\end{aligned}
$$
A: Hint:
Intuitively, $a_n$ gets as close as you want to $A$ (closer than $\epsilon$), for sufficiently large $N$.
Then the average will include a finite number of initial terms yielding a finite sum, and an infinite number of terms close to $A$. So in the limit, the average reduces to the close terms.
The proof needs to show more rigorously that even by increasing $N$ to support smaller $\epsilon$, the initial terms remain negligible.
