Question

The value of $\sum _{r=0}^{\infty }{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)$ is equal to

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

${\sum }_{r=0}^{\infty }{\tan }^{-1}\frac{(r+1)-r}{1+(r+1)\left(r\right)}={\sum }_{r=0}^{\infty }\left({\tan }^{-1}\right(r+1)-{\tan }^{-1}(r\left)\right)={\tan }^{-1}1-{\tan }^{-1}0+{\tan }^{-1}2-{\tan }^{-1}\mathrm{1.....}{\tan }^{-1}\infty$

$={\tan }^{-1}\infty -{\tan }^{-1}0=\frac{\pi }{2}$