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#### lfdahl

##### Well-known member

- Nov 26, 2013

- 719

\[\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ),\;\;\;\;\; k \geq 1,\]

- and deduce that

\[ \sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}.\]