Question

The value of $\sum _{r=1}^{\infty }{\tan }^{-1}\frac{2r}{2+r^2+r^4}$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is A $\frac{\pi }{4}$

${\tan }^{-1}\frac{2r}{r^4+r^2+2}={\tan }^{-1}\frac{2r}{1+(r^2+r+1)(r^2-r+1)}={\tan }^{-1}(r^2+r+1)-{\tan }^{-1}(r^2-r+1)$
$\underset{n\to \infty }{lim}\sum _{r=1}^n{\tan }^{-1}\frac{2r}{r^4+r^2+2}={lim}_{n\to \infty }({\tan }^{-1}3-{\tan }^{-1}+{\tan }^{-1}7-{\tan }^{-1}3+...{\tan }^{-1}x)$
$=\underset{n\to \infty }{lim}\left({\tan }^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)\right)=\underset{n\to \infty }{lim}\left({\tan }^{-1}\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n}+\frac{2}{n^2}}\right)\right)$
$={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$