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Math Help - alternating series sum

  1. #1
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    alternating series sum

    calculate the sum of (-1)^n (n/8^n)

    i know fourier coefficients and bessels inequality how to evaluate this
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  2. #2
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    Re: alternating series sum

    The sum, \sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}, can be evaluated by using only elementary techniques.

    Let S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}...(1)

    8S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^{n-1}}...(2)

    Adding (1) and (2) gives the incredible result

    9S=-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}=-1+\frac{1}{9}
    =\frac{-8}{9}

    so S=\frac{-8}{81}
    Last edited by sbhatnagar; November 24th 2012 at 03:42 AM.
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  3. #3
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    Re: alternating series sum

    i did not understand the step which produces the incredible result can yu please explain
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  4. #4
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    Re: alternating series sum

    Quote Originally Posted by prasum View Post
    i did not understand the step which produces the incredible result can yu please explain
    8S=-1+\frac{2}{8}-\frac{3}{8^2}+\cdots

    S=-\frac{1}{8}+\frac{2}{8^2}-\frac{3}{8^3}+\cdots
    _______________________________
    add them...
    9S=-1+\frac{1}{8}-\frac{1}{8^2}+\frac{1}{8^3}-\cdots

    =-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}
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  5. #5
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    Re: alternating series sum

    Quote Originally Posted by prasum View Post
    i did not understand the step which produces the incredible result can yu please explain
    You posted this in the calculus forum. So I assume it should be done using calculus.

    If y = \sum\limits_{k = 1}^\infty  {r^k }  = \frac{r}{{1 - r}},\;\left| r \right| < 1 then y' = \sum\limits_{k = 1}^\infty  {kr^{k - 1} }  = \frac{1}{{\left( {1 - r} \right)^2 }}.

    The multiply by r to get \sum\limits_{k = 1}^\infty  {kr^{k} }  = \frac{r}{{\left( {1 - r} \right)^2 }}.

    In this question r=\frac{-1}{8}.
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