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Math Help - Sum to infinity of series

  1. #1
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    Sum to infinity of series

    Find the sum of the following series:
    \frac{2}{1}(\frac{1}{3})+\frac{2}{3}(\frac{1}{3})^  3+\frac{2}{5}(\frac{1}{3})^5+...+\frac{2}{2r-1}(\frac{1}{3})^{2r-1}+...
    I don't know where to begin in this, can't recognize any of the known series.

    1-\frac{2}{4}+\frac{2.4}{4.8}-\frac{2.4.6}{4.8.12}-...
    My attempt:
    1-\frac{2}{1}(\frac{1}{4})+\frac{2.4}{1.2}(\frac{1}{  4})^2-\frac{2.4.6}{1.2.3}(\frac{1}{4})^3-...
    General term: \frac{(1)(2)(3)...(r)}{r!}(-\frac{1}{2})^r
    Which is the general term of the series (1+x)^n where n=1 and x=\frac{1}{2}
    So the sum =(1+(-\frac{1}{2}))^1=\frac{1}{2}
    Answer is: \frac{2}{3}

    Thanks!
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  2. #2
    Behold, the power of SARDINES!
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    Consider the series

    \displaystyle f(x)=2\sum_{n=0}^{\infty}{x}^{2n-2}=\frac{2}{x^2(1-x^2)}

    Now integrate to get

    \displaystyle g(x)=\int f(x)dx=2\sum_{n=0}^{\infty}\frac{1}{2n-1}{x}^{2n-1}=\int\frac{2}{x^2(1-x^2)}dx

    Note that this is your series when x=\frac{1}{3}
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  3. #3
    Junior Member Renji Rodrigo's Avatar
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    Other solution ( using D for derivative and arctgh(x) for ArcTanh(x) )

    we know that  D arctgh (x) =\frac{1}{1-x^{2}} and  arctgh (0)=0 so

     \int^{x}_{0}\frac{1}{1-y^{2}}dy=arctgh(x)

    using the geometric series for \frac{1}{1-y^{2}} we have
     \int^{x}_{0}\frac{1}{1-y^{2}}dy=\int^{x}_{0}\sum\limits^{\infty}_{k=0}y^{  2k}dy=   \sum\limits^{\infty}_{k=0}\int^{x}_{0}y^{2k}dy=\su  m\limits^{\infty}_{k=0}\frac{x^{2k+1}}{2k+1}

    so  arctgh (x)=\sum\limits^{\infty}_{k=0}\frac{x^{2k+1}}{2k+1  }.
    taking x=1/3


     2arctgh (\frac{1}{3})=2\sum\limits^{\infty}_{k=0}\frac{ (1/3)^{2k+1}}{2k+1}.
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  4. #4
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    So the sum would be \int\frac{2}{x^2(1-x^2)}dx when x=\frac{1}{3}?
    \int^{\frac{1}{3}}_0 \frac{2}{x^2(1-x^2)}dx=[-\frac{2}{x}-\ln(1-x)+\ln(x+1)]^{\frac{1}{3}}_0=-5.31 to three s.f.
    answer is \ln 2
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  5. #5
    Junior Member Renji Rodrigo's Avatar
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    Another try
     \sum\limits^{\infty}_{k=0}\frac{1}{a^{2k}(2k+1)}=\  sum\limits^{\infty}_{k=0}\frac{1}{a^{2k}}\int^{1}_  {0}x^{2k}=\int^{1}_{0}\sum\limits^{\infty}_{k=0}\b  igg(\frac{x^{2}}{a^{2}}\bigg)^{k}dx=  \int^{1}_{0}\frac{a^{2}}{a^{2}-x^{2}}dx

    by the parcial fraction decomposition
     \frac{a^{2}}{a^{2}-x^{2}}=\frac{a}{2}\bigg(\frac{1}{a+x}-\frac{1}{x-a} \bigg)


    we get
     2\sum\limits^{\infty}_{k=0}\frac{1}{a^{2k+1}(2k+1)  }=\ln (\frac{a+1}{a-1})
    taking a=3

    2\sum\limits^{\infty}_{k=0}\frac{1}{3^{2k+1}(2k+1)  }=\ln (\frac{4}{2})= \ln (2)
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  6. #6
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    Opalg's Avatar
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    Quote Originally Posted by arze View Post
    1-\frac{2}{4}+\frac{2.4}{4.8}-\frac{2.4.6}{4.8.12}-...
    My attempt:
    1-\frac{2}{1}(\frac{1}{4})+\frac{2.4}{1.2}(\frac{1}{  4})^2-\frac{2.4.6}{1.2.3}(\frac{1}{4})^3-...
    General term: \frac{(1)(2)(3)...(r)}{r!}(-\frac{1}{2})^r (Correct.)
    Which is the general term of the series (1+x)^n where n=1 (Should be n = 1.) and x=\frac{1}{2}
    So the sum =(1+(-\frac{1}{2}))^1=\frac{1}{2} (Make that (1+\frac{1}{2})^{-1}=\frac23 .)
    Answer is: \frac{2}{3}
    ..
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