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Math Help - An infinite sum

  1. #1
    Junior Member gusztav's Avatar
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    An infinite sum

    Hello,

    I'd be really grateful for any help with finding the following sum:

    \sum_{n=1}^{\infty} \arctan \frac{1}{2n^2}

    Many thanks!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by gusztav View Post
    Hello,

    I'd be really grateful for any help with finding the following sum:

    \sum_{n=1}^{\infty} \arctan \frac{1}{2n^2}

    Many thanks!
    \frac{1}{2n^2}=\frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)} and thus \tan^{-1} \left(\frac{1}{2n^2} \right)=\tan^{-1}(2n+1) - \tan^{-1}(2n-1). now you have a telescoping sum with the value equal to \lim_{n\to\infty}(\tan^{-1}(2n+1) - \tan^{-1}1)=\frac{\pi}{4}.
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