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Math Help - Express as a telescoping series. If convergent find its sum.

  1. #1
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    Express as a telescoping series. If convergent find its sum.

    \sum\limits_{n=1}^{\infty} \frac{1}{n(n+2)}

    I used partial fractions and got this term:

    \sum\limits_{n=1}^{\infty} \frac{1}{n(n+2)} = \frac{\frac{1}{2}}{n} - \frac{\frac{1}{2}}{n+2}

    Then factored out the 1/2 and I'm began to expand,

    \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+2} = ....

    After taking the limit as n goes to infinity I eventually get that the answer converges to 3/4. Can someone please confirm?
    Last edited by superduper1; November 18th 2012 at 09:24 PM.
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  2. #2
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    Re: Express as a telescoping series. If convergent find its sum.

    This is why the series is called "telescoping":

    \frac{1}{2}\sum\limits_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) =

    \frac{1}{2}\left(\sum\limits_{n=1}^{\infty} \frac{1}{n} - \sum\limits_{n=1}^{\infty}\frac{1}{n+2} \right)=

    \frac{1}{2}\left(\sum\limits_{n=1}^{\infty} \frac{1}{n} - \sum\limits_{n=3}^{\infty}\frac{1}{n} \right)=

    \frac{1}{2}\left( \frac{1}{1} + \frac{1}{2} \right)

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