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Proof of 2nd Derivative of a Sum of a Geometric Series

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    I am trying to prove how \(g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}\).

    I don't know what I am doing wrong and am at my wits end.

    3. The attempt at a solution (The index of the summation is always k=2 to infinity)

    Ʃak(k-1)r^(k-1)
    =a Ʃ k(k-1)r^(k-2)
    =a Ʃ (r^k)''
    =a (r^2 / (1-r)''

    From this point I get a mess, and the incorrect answer. The thing I have a problem is I think Ʃ (r^k)'' when k is from 2 to infinity is r^2/(1-r), since the first term in this sequence is r^2. I don't think that is correct though
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 8, 2012 #2

    SammyS

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    To fix your LaTeX expression, I inserted the correct tags, itex and /itex in [ ] brackets.

    Since the sum starts at k=2, you would not have those first two zero terms. It would be:

    [itex]g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}[/itex]

    So, what is it that you're supposed to be proving ?
     
  4. Nov 8, 2012 #3
    it turns out that what i was doing was actually correct
     
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