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)

$\frac{\pi }{2}$

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

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

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

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

$={\tan }^{-1}1-{\tan }^{-1}0+{\tan }^{-1}2-{\tan }^{-1}1+...\dots +{\tan }^{-1}\infty$

$={\tan }^{-1}\infty -{\tan }^{-1}0$

$=\frac{\pi }{2}-0$

$=\frac{\pi }{2}$.