Question

The sum of the series upto infinity
${\tan }^{-1}(\frac{1}{{2.1}^2})+{\tan }^{-1}(\frac{1}{{2.2}^2})+{\tan }^{-1}(\frac{1}{{2.3}^2})+...\infty$ is

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

Answer 1

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

$\sum _{r=1}^{\infty }{\tan }^{-1}(\frac{1}{2r^2})=\sum _{r=1}^{\infty }{\tan }^{-1}(\frac{2}{1+4r^2-1})=\sum _{r=1}^{\infty }{\tan }^{-1}(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)})=\sum _{r=1}^{\infty }{\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1))$
$=\underset{n\to \infty }{lim}({\tan }^{-1}3-{\tan }^{-1}1)+({\tan }^{-1}5-{\tan }^{-1}3)+....+\left({\tan }^{-1}\right(2n+1)-{\tan }^{-1}(2n-1)$
$=\underset{n\to \infty }{lim}\left({\tan }^{-1}\right(2n+1)-{\tan }^{-1}1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$