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Thread: simple sigma problem

  1. #1
    Sep 2008

    Wink simple sigma problem

    No. 20 / Thank you for helping me
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
    May 2008
    Chicago, IL
    Quote Originally Posted by narbe View Post

    No. 20 / Thank you for helping me
    Note that $\displaystyle \sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^n \left[\frac{1}{k}-\frac{1}{k-1}\right]$ is a telescoping series. So when you go to expand it, we are left with:

    $\displaystyle \underbrace{1-\frac{1}{2}}_{k=1}+\underbrace{\frac{1}{2}-\frac{1}{3}}_{k=2}+\underbrace{\frac{1}{3}-\frac{1}{4}}_{k=3}+\dots+\underbrace{\frac{1}{n}-\frac{1}{n+1}}_{k=n}$

    So we see that all the terms cancel out, except for the first and last terms.

    So $\displaystyle \sum_{k=1}^n \left[\frac{1}{k}-\frac{1}{k-1}\right]=\color{red}\boxed{1-\frac{1}{n+1}}$

    Does this make sense?

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