The sum of the infinite series

$$\cot^{-1}2 +\cot^{-1}8 +\cot^{-1}18 +\cot^{-1}32 +\cot^{-1}50 + ......$$ is

We see that 2,8,18... can be represented as $$2n^2$$ where n$$\in\ N$$

Required sum = $$\Sigma_1^{\infty}\left(\cot^{-1}\left(2n^2\ \right)\right)\ =\ \Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ 1}{2n^2}\ \right)\right)$$

=$$\Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ 2}{4n^2}\ \right)\right)$$

= $$\Sigma_1^{\infty}\left(\tan^{-1}\left(\frac{\ \left(2n+1\right)-\left(2n-1\right)}{1+\left(4n^2-1\right)}\ \right)\right)$$

= $$\Sigma\ _1^{\infty\ }\ \tan^{-1}\left(2n+1\right)\ -\ \tan^{-1}\left(2n-1\right)$$

= $$\tan^{-1}\infty\ -\ \tan^{-1}1$$

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