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Cauchy geometric seq. proof

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose the sequence (Sn) is defined as:
    |Sn+1-Sn|<2-n
    show that this is a cauchy sequence

    2. Relevant equations
    hint: prove the polygon identity such that
    d(Sn,Sm)≤d(Sn,Sn+1)+d(Sn+1,Sn+2)...+d(Sm-1,Sm)

    3. The attempt at a solution
    I have defined Sm and Sn and created the inequality that:
    d(Sn,Sm)=|2-n-2-m|>||Sn+1-Sn|-|Sm+1-Sm||
     
  2. jcsd
  3. May 28, 2012 #2
    Using the hint, you'd get |Sn-Sm|<|Sn-Sn+1|+...+|Sm-1-Sm|<2-n+...+2-m
    Is there any type of cauchy sequence (perhaps of partial sums) that would fit the latter half?
     
  4. May 28, 2012 #3
    ok I think I get what I can do now with the polygon ineq. except I thought that it would only be less than 2-n+...+2m-1also I don't know if I need to prove the polygon identity or not. if I did how would I go about it? would I use a fact of geometric series?
     
  5. May 29, 2012 #4
    Proving the "polygon" identity comes as a direct result of the definition of a metric, and the triangle inequality. A simple proof by induction would suffice.
    While it is true you could remove the factor of 2-m from the inequality, you can add it on and still have the inequality hold true (plus it's better looking notation wise).
    We know Ʃ2-n is a geometric series, and thus converges. Convergent sequences of partial sums (partial sums of a geometric series) are cauchy sequences. So Ʃ2-k from 1 to n minus Ʃ2-k from 1 to m would be less than ε
     
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