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Math Help - Help with a proof (Convergence)

  1. #1
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    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!
    Attached Thumbnails Attached Thumbnails Help with a proof (Convergence)-exercise.png  
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  2. #2
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    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.
    Last edited by romsek; November 29th 2013 at 12:54 PM.
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  3. #3
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    Re: Help with a proof (Convergence)

    Quote Originally Posted by davidciprut View Post
    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<N 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
    Last edited by Plato; November 29th 2013 at 12:25 PM.
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