Question

Find the value of $\sum _{r=0}^{\infty }{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $3\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{2}$

The value of ${\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)={\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)$
$={\tan }^{-1}(r+1)-{\tan }^{-1}\left(r\right)$
$\Rightarrow$ $\sum _{r=0}^n[{\tan }^{-1}(r+1)-{\tan }^{-1}\left(r\right)]$
$={\tan }^{-1}(n+1)-{\tan }^{-1}\left(0\right)$
$={\tan }^{-1}(n+1)$
$\Rightarrow$ $\sum _{r=0}^{\infty }{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)={\tan }^{-1}\left(\infty \right)=\frac{\pi }{2}$