Question

The sum of the infinite series
$\cot^{-1}(2\cdot 1^2)+\cot^{-1}(2\cdot 2^2)+\cot^{-1}(2\cdot 3^2)+\cot^{-1}(2\cdot 4^2)+.......$ is equal to

• A. $\frac {\pi}{5}$
• B. $\frac {\pi}{4}$
• C. $\frac {\pi}{3}$
• D. $\frac {\pi}{2}$

Answer 1

Answer: (B)

$S = cot^{-1} 2\cdot1^{2} + cot^{-1} 2\cdot2^{2} $
$+cot^{-1} 2\cdot3^{2} +cot^{-1}2\cdot4^{2}+.....$
$ T_{n} = cot^{-1}\left(2\cdot n^{2}\right) $
$ = tan^{-1} \frac{1}{2n^{2}} = tan^{-1}\left(\frac{2}{4n^{2}}\right)$
$= tan^{-1}\left[\frac{\left(2n+1\right)-\left(2n-1\right)}{1+\left(4n^{2}-1\right)}\right] $
$ = tan^{-1} \left(2n+1\right) - tan^{-1}\left(2n-1\right) $
$\therefore S = tan^{-1} 3 - tan^{-1}1 +tan^{-1}5 - tan^{-1} 3$
$+tan^{-1}7+tan^{-1}5+.....\infty $
$ = tan^{-1} \infty -tan^{-1}1 $
$ =\frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$