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Find the sum of this series

  1. Apr 23, 2009 #1

    Σ (x2n-1)/(2n-1)
    n=1

    what i tried to do was take the taylor series:


    Σ (xn)=1/(1-x)
    n=1

    so i can substitute x2 for x


    Σ (x2n)=1/(1-x2)
    n=1


    now i need to to use some form of integration/derivative to get to my series from the taylor series,
    my problem is that integral will give me x2n+1/(2n+1) , and derivative will give me 2n*x2n-1, but i need the format of the integration- a division format- but the numberss from the derivative- 2n-1 -
     
  2. jcsd
  3. Apr 23, 2009 #2

    dx

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    Gold Member

    Hint: Change the limits to n = 0 to ∞ and (2n - 1) to (2n + 1).
     
  4. Apr 23, 2009 #3
    tell me if this is right


    Σ x2n-1/(2n-1)
    n=1
    =

    Σ x2(n+1)-1/(2(n+1)-1)
    n=0
    =

    Σ x2n+1/(2n+1)
    n=0

    fx= 1/(1-x) = Σ xn
    x==>x2
    fx= 1/(1-x2) = Σ x2n

    [tex]\int[/tex]fx=[tex]\int[/tex]1/(1-x2)=[tex]\int[/tex]Σ x2n

    0.5*ln|(1+x)/(1-x)| = x2n+1/2n+1

    and i can do all of this because i say n=1, and n*=0, therefore n* goes from 0-∞ and n goes from 1-∞, making n*=n+1 so i can use the same taylor series, but instead of n, i write n+1
     
  5. Apr 23, 2009 #4

    dx

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    Homework Helper
    Gold Member

    Ya looks good.
     
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