Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

Given, $S=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{2n}{n^4+n^2+2} ...\left(\mathrm{i}\right)$

Let $n^4+n^2+1=\left[{\left(n^2\right)}^2+1^2+2\left(n^2\right)\left(1\right)\right]-n^2$

$={\left(n^2+1\right)}^2-n^2$

$=\left(n^2+n+1\right)\left(n^2-n+1\right) ...\left(\mathrm{ii}\right)$

Let $u_n={\tan }^{-1}\left(\frac{2n}{n^4+n^2+2}\right)$ 

$={\tan }^{-1}\frac{2n}{1+\left(n^4+n^2+1\right)}$

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

$u_n={\tan }^{-1}\left(n^2+n+1\right)-{\tan }^{-1}\left(n^2-n+1\right) ...\left(\mathrm{iii}\right)$

On putting $n=1, 2, 3,\dots \dots$ successively in Eq. (iii), we get

$u_1={\tan }^{-1}3-{\tan }^{-1}1$

$u_2={\tan }^{-1}7-{\tan }^{-1}3$

$u_3={\tan }^{-1}13-{\tan }^{-1}7$

$u_n={\tan }^{-1}\left(n^2+n+1\right)-{\tan }^{-1}\left(n^2-n+1\right)$

On adding vertically, we get

$\sum _{n=1}^{\infty }{u}_n={\tan }^{-1}\left(n^2+n+1\right)-{\tan }^{-1}1$

$S=\underset{n\to \infty }{\lim }\sum _{n=1}^{\infty }{u}_n$   [from Eq. (i)]

$=\underset{n\to \infty }{\lim }{\tan }^{-1}\left(n^2+n+1\right)-{\tan }^{-1}1$

$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$