I uploaded a picture, a proof of a statement. There are things that I didn't understand. I translated the question from Hebrew to English so I hope it's clear enough. Thanks in advance!
This proof splits the sequence x of length n into two subsequences, the first M elements, and the rest.
The property of elements m+1 through n is that |x_{n}| < $\displaystyle \epsilon$/2 for n > m
Now given that the sum of the first m+1 elements is finite you can always find some N such that the above some divided by N is also less than $\displaystyle \epsilon$/2.
It then composes the magnitude of the average of the n elements of the sequence into the magnitude of the (sum of first m elements + sum of last n-m elements)/n
Using triangle inequality it breaks that up into the
(sum of the magnitudes of first m elements of x)/n + (sum of magnitudes of last (n-m) elements)/n
Now you know by construction that the last (n-m) elements magnitudes are all < $\displaystyle \epsilon$/2
So the sum of them divided by n is < (n-m)/n $\displaystyle \epsilon$/2 (do you understand this? there are n-m elements in the sum)
Also by construction we know the sum of the first m elements divided by n is < $\displaystyle \epsilon$/2.
Combine these and you get the full sum of n magnitudes of x < $\displaystyle \epsilon$/2 + (n-m)/n $\displaystyle \epsilon$/2 < $\displaystyle \epsilon$
So in sum what this does is for a given $\displaystyle \epsilon$/2 it finds the element M where the magnitude of x becomes less than $\displaystyle \epsilon$/2.
It then takes more elements of the sequence, up to x_{N} so that the sum of the first M elements divided by N < $\displaystyle \epsilon$/2. This can be done because the sum of the first M elements is finite so there must be some N big enough.
The (sum of first M elements)/N < $\displaystyle \epsilon$/2 because we chose N large enough
The (sum of the last N-M elements)/N < $\displaystyle \epsilon$/2 because we chose M large enough.
See if this helps.
Surely you have made some mistakes in the statement?
It should be $\displaystyle {A_N} = \sum\limits_{k = 1}^N {\frac{{{x_k}}}{N}} $ otherwise it is a trivial question.
Note that if $\displaystyle M<N$ then $\displaystyle {A_N} = {A_M} + ({A_N} - {A_M}) = \sum\limits_{k = 1}^M {\frac{{{x_k}}}{N}} + \sum\limits_{k = M + 1}^N {\frac{{{x_k}}}{N}} $
If $\displaystyle M$ is fixed, then so is the sum $\displaystyle A_M$. So we can make $\displaystyle \frac{|A_M|}{N}\le\frac{\epsilon }{2}$.
That is part of it.
In the sum $\displaystyle A_N$ there are $\displaystyle N$ terms so is there are $\displaystyle N-M$ terms is $\displaystyle A_N-A_M$.
Note that $\displaystyle \frac{N-M}{N}=1-\frac{M}{N}<1$