# Thread: Help with a proof (Convergence)

1. ## Help with a proof (Convergence)

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!

2. ## Re: Help with a proof (Convergence)

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 |xn| < $\epsilon$/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 $\epsilon$/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 < $\epsilon$/2
So the sum of them divided by n is < (n-m)/n $\epsilon$/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 < $\epsilon$/2.

Combine these and you get the full sum of n magnitudes of x < $\epsilon$/2 + (n-m)/n $\epsilon$/2 < $\epsilon$

So in sum what this does is for a given $\epsilon$/2 it finds the element M where the magnitude of x becomes less than $\epsilon$/2.

It then takes more elements of the sequence, up to xN so that the sum of the first M elements divided by N < $\epsilon$/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 < $\epsilon$/2 because we chose N large enough
The (sum of the last N-M elements)/N < $\epsilon$/2 because we chose M large enough.

See if this helps.

3. ## Re: Help with a proof (Convergence)

Originally Posted by davidciprut
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!
Surely you have made some mistakes in the statement?
It should be ${A_N} = \sum\limits_{k = 1}^N {\frac{{{x_k}}}{N}}$ otherwise it is a trivial question.

Note that if $M then ${A_N} = {A_M} + ({A_N} - {A_M}) = \sum\limits_{k = 1}^M {\frac{{{x_k}}}{N}} + \sum\limits_{k = M + 1}^N {\frac{{{x_k}}}{N}}$

If $M$ is fixed, then so is the sum $A_M$. So we can make $\frac{|A_M|}{N}\le\frac{\epsilon }{2}$.
That is part of it.

In the sum $A_N$ there are $N$ terms so is there are $N-M$ terms is $A_N-A_M$.

Note that $\frac{N-M}{N}=1-\frac{M}{N}<1$