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  1. #1
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    series

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


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

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

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


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

    It should be,
    \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.

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

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

    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}

    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,
    S_n = \frac{3}{1} + \frac{3}{2} - \frac{3}{n+1} - \frac{3}{n+2}

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