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Sum of infinite series

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find the sum of a given infinite series when [itex]|x|<1[/itex] (which is the radius of this series)

    2. Relevant equations

    [itex]\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n+1}}{4n^2-1} [/itex]

    3. The attempt at a solution
    I've tried to do the following:

    [itex]S'(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n}}{2n-1} \\
    S''(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{2nx^{2n-1}}{2n-1}\\
    S'''(x) = 2\sum_{n=1}^{∞}(-1)^{n+1}nx^{2(n-1)}\\[/itex]


    And I was thinking about substitution [itex]t = x^2[/itex], but I had no success.
     
  2. jcsd
  3. Jun 4, 2012 #2

    tiny-tim

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    hi stefaneli! :smile:
    [itex]= x\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}[/itex] :wink:
     
  4. Jun 4, 2012 #3
    Thanks tiny-tim. I don't know how I haven't noticed. Stupid.:) But can it be done the way I started? Just curious.
     
  5. Jun 4, 2012 #4

    tiny-tim

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    i suppose so, but you'd need to be very careful when you change the variable :frown:
     
  6. Jun 4, 2012 #5
    Try doing partial fractions. IE 1/(4n^2-1)= 1/2*(1/(2n-1)-1/(2n+1))
     
  7. Jun 4, 2012 #6
    Your answer will be something like 1/2*((x^2-1)arctan(x)-x), just use partial fractions, and get it into a more manageable form
     
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