Question

The value of ${\sum }_{n=1}^{\infty }[{\tan }^{-1}\left[\frac{2}{n+2}\right]-{\tan }^{-1}\left[\frac{n-1}{n+2}\right]]$ is equal to-

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

Answer 1

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

${\sum }_{n=1}^{\infty }\left[{\tan }^{-1}\right[\frac{2}{n+2}]-{\tan }^{-1}[\frac{n-1}{n+2}\left]\right]$

$\Rightarrow {\sum }_{n=1}^{\infty }{\tan }^{-1}\left[\frac{(3-n)(n+2)}{n^2+4+4n-2n+2}\right]$

$\Rightarrow {\tan }^{-1}\frac{(\frac{2}{n+2}-\frac{n-1}{n+2})}{1-\left(\frac{2}{n+2}\right)\left(\frac{n-1}{n+2}\right)}$

$\Rightarrow {\sum }_{n=1}^{\infty }{\tan }^{-1}\left[\frac{(3-n)(n+2)}{n^2+6+2n}\right]$

$\Rightarrow {\tan }^{-1}\left(\frac{2\times 3}{9}\right)+{\tan }^{-1}\left(\frac{1\times 4}{14}\right)+{\tan }^{-1}\left(0\right)+..........+{\tan }^{-1}(-1)$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{2-n+1}{n+2}}{\frac{(n+2{)}^2-2(n-1)}{(n+2{)}^2}}\right]$

$\Rightarrow {\tan }^{-1}\left(\frac{2}{3}\right)+{\tan }^{-1}\left(\frac{2}{7}\right)+...........+{\tan }^{-1}\left(1\right)$

$\Rightarrow \frac{\pi }{4}$