Hello,
I'd be really grateful for any help with finding the following sum:
$\displaystyle \sum_{n=1}^{\infty} \arctan \frac{1}{2n^2}$
Many thanks!
$\displaystyle \frac{1}{2n^2}=\frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}$ and thus $\displaystyle \tan^{-1} \left(\frac{1}{2n^2} \right)=\tan^{-1}(2n+1) - \tan^{-1}(2n-1).$ now you have a telescoping sum with the value equal to $\displaystyle \lim_{n\to\infty}(\tan^{-1}(2n+1) - \tan^{-1}1)=\frac{\pi}{4}.$