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  1. #1
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    series

    $\displaystyle \sum_{n=1}^\infty \frac{6}{n(n+2)}$


    $\displaystyle \frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$

    $\displaystyle S_n = (6-\frac{6}{3}) + (\frac{6}{2}-\frac{6}{4}) + (\frac{6}{3}-\frac{6}{5})+$...$\displaystyle + (\frac{6}{n} - \frac{6}{n+2})$

    not sure what to do next
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  2. #2
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    Quote Originally Posted by viet View Post

    $\displaystyle \frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$

    [
    The above is not true.
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  3. #3
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    Quote Originally Posted by viet View Post
    $\displaystyle \sum_{n=1}^\infty \frac{6}{n(n+2)}$


    $\displaystyle \frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$
    First that is wrong.

    It should be,
    $\displaystyle \frac{6}{n(n+2)} = \frac{3}{n} - \frac{3}{n+2}$

    Now, here is how I explain it to people, look at the partial sums.

    $\displaystyle S_1= \frac{3}{1} - \frac{3}{3}$

    $\displaystyle S_2 = \frac{3}{1} - \frac{3}{3} + \frac{3}{2} - \frac{3}{4}$

    $\displaystyle S_3 = \frac{3}{1} - \frac{3}{3}+\frac{3}{2} - \frac{3}{4} + \frac{3}{3} - \frac{3}{5}= \frac{3}{1} + \frac{3}{2} - \frac{3}{4} - \frac{3}{5}$

    $\displaystyle S_4 = \frac{3}{1}+\frac{3}{2} - \frac{3}{4} - \frac{3}{5} + \frac{3}{4} - \frac{3}{6} = \frac{3}{1}+\frac{3}{2} - \frac{3}{5} - \frac{3}{6}$

    It seems the pattern is,
    $\displaystyle S_n = \frac{3}{1} + \frac{3}{2} - \frac{3}{n+1} - \frac{3}{n+2}$

    Now just see what happens as $\displaystyle n\to \mbox{Hacker}$.
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