$\text{The value of }\sum _{r=0}^{\mathrm{\infty }} {\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)\text{ =}\frac{\mathrm{\pi }}{k}\text{ then k=}$

${\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)={\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)$

$={\tan }^{-1}(r+1)-{\tan }^{-1}\left(r\right)$

$\begin{array}{l}Given question\Rightarrow \sum _{r=0}^n \left[{\tan }^{-1}(r+1)-{\tan }^{-1}\left(r\right)\right] \\ = ( {\tan }^{-1}1-{\tan }^{-1}0) + ({\tan }^{-1}2-{\tan }^{-1}1) + ........+({\tan }^{-1}\left(n+1\right) -{\tan }^{-1}(n) )\\ ={\tan }^{-1}(n+1)-{\tan }^{-1}\left(0\right)\\ ={\tan }^{-1}(n+1)\\ \Rightarrow \underset{n\to \infty }{\lim }\sum _{r=0}^{\mathrm{n}} {\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)={\tan }^{-1}\left(\mathrm{\infty }\right)=\frac{\pi }{2}\end{array}$