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Math Help - Compute the following sum.

  1. #1
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    Compute the following sum.

    The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

    Thanks.
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  2. #2
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    Quote Originally Posted by feyomi View Post
    The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

    Thanks.
    \sum_{i=1}^{\infty} (\frac{1}{3})^{n-1} = [\sum_{i=0}^{\infty} (\frac{1}{3})^{n-1}] - (\frac{1}{3})^{-1} = \frac{9}{2} - 3 = \frac{3}{2}

    Since,

    \sum_{i=0}^{\infty} (\frac{1}{3})^{n-1}=(\frac{1}{3})^{-1} \sum_{i=0}^{\infty} (\frac{1}{3})^{n} = \frac{(\frac{1}{3})^{-1}}{1 - \frac{1}{3}} = \frac{9}{2}
    Last edited by Anonymous1; May 6th 2010 at 01:49 PM.
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  3. #3
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    Quote Originally Posted by feyomi View Post
    The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

    Thanks.
    Hi feyomi,

    S_{\infty}=(1)\left(\frac{1}{3}\right)^0+(2)\left(  \frac{1}{3}\right)^1+(3)\left(\frac{1}{3}\right)^2  +....

    =(1)+(2)\left(\frac{1}{3}\right)^1+(3)\left(\frac{  1}{3}\right)^2+....

    \frac{1}{3}S_{\infty}=\frac{1}{3}+(2)\left(\frac{1  }{3}\right)^2+(3)\left(\frac{1}{3}\right)^3+....

    S_{\infty}-\frac{1}{3}S_{\infty}=1+(2-1)\left(\frac{1}{3}\right)+(3-2)\left(\frac{1}{3}\right)^2+(4-3)\left(\frac{1}{3}\right)^3+.....

    =1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\f  rac{1}{3}\right)^3+....

    =geometric\ series

    For the geometric series, the first term is 1 and the common ratio is 1/3,
    hence it's sum to infinity is

    \frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}

    \frac{2}{3}S_{\infty}=\frac{3}{2}

    S_{\infty}=\frac{3}{2}\frac{3}{2}=\frac{9}{4}
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