$A_n$=$\frac {a_1+a_2+...a_n}{n}$ is monotonic if $a_n$ is monotonic if $a_n$ is monotonic increasing/decreasing show that sequence 
$A_n$=$\frac {a_1+a_2+...a_n}{n}$ is also monotonic increasing/decreasing.
my attempt:
I intially thought of using induction since $A_2>A_1$ when $a_2>a_1$ so base case is available. but to prove $A_{n+1}>A_n$ doesnt show up easy. Any other way?
 A: Proving the generalized case is very similar to the base case, because you can write $A_{n+1} = \frac{n}{n+1}A_n + \frac{1}{n+1} a_{n+1}$, which looks very similar to $A_2 = \frac{1}{2}A_1 + \frac{1}{2} a_{2}$
Basically, it amounts to stating why the relative contribution from $a_{n+1}$ is at least as large as the relative contribution from any of the previous elements $a_j, j<n$ (the reason for which is monotonicity of the sequence $(a_j)_{j>1}$).
A: We can also solve this problem by using Stolz–Cesàro theorem.
$$\lim\limits_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$
$$\lim\limits_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim\limits_{n\to\infty}\frac{a_n}{b_n}$$ If $b_n$ is strictly monotone and divergent sequence.
So we $A_n$ can be written as 
$$\lim\limits_{n\to\infty}\frac{(a_1+a_2+\dots+a_n+a_{n+1})-(a_1+a_2+\dots+a_n)}{n+1-n}$$
$$A_n=\lim\limits_{n\to\infty}\frac{a_{n+1}}{1}$$
if $a_n$ is monotonic increasing/decreasing we show that sequence $A_n$ is also monotonic increasing/decreasing.
