An infinite sum

**gusztav** · June 28th 2009, 05:57 AM

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!

**NonCommAlg** · June 28th 2009, 06:04 AM

Originally Posted by gusztav

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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