- #1

lfdahl

Gold Member

MHB

- 749

- 0

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

- and deduce that

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