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Math Help - Sum of the series

  1. #1
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    Sum of the series

    I am struggling to see a way of finding the infinite sum of the series

    2sin(\frac{1}{k}-\frac{1}{k+1})cos(\frac{1}{k}+\frac{1}{k+1})

    sin(\frac{1}{k}-\frac{1}{k+1}) converges to 0 and cos(\frac{1}{k}+\frac{1}{k+1}) converges to 1.

    Expanding and simplifying the trig functions just yields
    2(sin(\frac{1}{k})cos(\frac{1}{k})cos(\frac{2}{k+1  })-sin(\frac{1}{k+1})cos(\frac{1}{k+1})cos(\frac{2}{k  }))

    Which is not of any use

    We didn't cover many methods for finding the sum of series in class, just ways of showing that a series converges
    Could someone point me in the right direction?
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Sum of the series

    2\sin\left(\frac1k - \frac1{k+1}\right)\cos\left(\frac1k + \frac1{k+1}\right)\,=\,\sin\frac2k-\sin\frac2{k+1}

    \sum_{k=1}^n\left[\sin\frac2k-\sin\frac2{k+1}\right]\,=\,\sin2-\sin\frac2{n+1}\to\sin2\ \mbox{as}\ n\to\infty
    Thanks from Shakarri
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  3. #3
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    Re: Sum of the series

    At first I didnt see what you did but now I see its a telescoping series! thank you
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