The sum of the the series ${\cot }^{-1}2+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32+$…….. is

The given series

$={\cot }^{-1}2(1{)}^2+{\cot }^{2}2(2{)}^2+{\cot }^{-1}2(3{)}^2+{\cot }^{-1}2(4{)}^2+\dots$

$\begin{array}{l}\therefore {\mathrm{T}}_{\mathrm{n}}={\cot }^{-1}\left(2{\mathrm{n}}^2\right)={\tan }^{-1}\frac{1}{2{\mathrm{n}}^2}\\ ={\tan }^{-1}\left[\frac{(2\mathrm{n}+1)-(2\mathrm{n}-1)}{1+(2\mathrm{n}+1)(2\mathrm{n}-1)}\right]\\ ={\tan }^{-1}(2\mathrm{n}+1)-{\tan }^{-1}(2\mathrm{n}-1)\\ \therefore {\mathrm{T}}_1={\tan }^{-1}3-{\tan }^{-1}1,{\mathrm{T}}_2={\tan }^{-1}5-{\tan }^{-1}3,\\ {\mathrm{T}}_3={\tan }^{-1}7-{\tan }^{-1}5\end{array}$

and so on.

$\begin{array}{l}\therefore {\mathrm{S}}_{\mathrm{n}}={\tan }^{-1}(2\mathrm{n}+1)-{\tan }^{-1}1\\ \therefore \underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\mathrm{S}}_{\mathrm{n}}={\tan }^{-1}\mathrm{\infty }-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{4}\end{array}$