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Thread: Sum to infinity of series

  1. #1
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    Sum to infinity of series

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

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

    Thanks!
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  2. #2
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    Consider the series

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

    Now integrate to get

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

    Note that this is your series when $\displaystyle x=\frac{1}{3}$
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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 $\displaystyle D arctgh (x) =\frac{1}{1-x^{2}}$ and $\displaystyle arctgh (0)=0 $ so

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

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

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


    $\displaystyle 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 $\displaystyle \int\frac{2}{x^2(1-x^2)}dx$ when $\displaystyle x=\frac{1}{3}$?
    $\displaystyle \int^{\frac{1}{3}}_0 \frac{2}{x^2(1-x^2)}dx=[-\frac{2}{x}-\ln(1-x)+\ln(x+1)]^{\frac{1}{3}}_0=-5.31$ to three s.f.
    answer is $\displaystyle \ln 2$
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  5. #5
    Junior Member Renji Rodrigo's Avatar
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    Another try
    $\displaystyle \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
    $\displaystyle \frac{a^{2}}{a^{2}-x^{2}}=\frac{a}{2}\bigg(\frac{1}{a+x}-\frac{1}{x-a} \bigg)$


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

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