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Math Help - Summation

  1. #1
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    Summation

    Use the identity \displaystyle \frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1} to derive a formula for \displaystyle \sum_{i=1}^{n}\frac{1}{i(i+1)}.

    I am not sure how to do this.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Develop the sum for n=4... what do you see?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    \displaystyle \sum_{i=1}^{n}\frac{1}{i(i+1)}=\displaystyle \sum_{i=1}^{n}\frac{1}{i}-\frac{1}{i+1}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n-1}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{n+1}

    By the way it's very famous series!

    Telescoping series - Wikipedia, the free encyclopedia
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