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Math Help - The sum of another problematic sequence.

  1. #1
    Super Member Showcase_22's Avatar
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    The sum of another problematic sequence.



    That's the question and my working so far. I can't really see what the next step is or if this method will work out at the end.

    Can someone please help me?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post


    That's the question and my working so far. I can't really see what the next step is or if this method will work out at the end.

    Can someone please help me?
    Go back to the original series and substitute \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} ....
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