What is $x$, if $\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots = 1$? 
Let
$$S_{n}=\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots \quad\text{($n$ terms)}$$
where $x>0$. If $\lim _{n \to \infty} S_{n}=1$, then  find the value of $x$.

Can the given series be converted to telescopic series? I converted in into $\tan^{-1}$ but in every term $x$ is in numerator? Could someone please given some hint?

We have 
\begin{eqnarray*}
\cot^{-1}(A)-\cot^{-1}(B)=\cot^{-1}\left(\frac{AB+1}{B-A} \right).
\end{eqnarray*}
In your case this gives
\begin{eqnarray*}
\cot^{-1}\left(\frac{i(i+1)x}{2}+\frac{2}{x} \right)=\cot^{-1}\left(\frac{ix}{2} \right)-\cot^{-1}\left(\frac{(i+1)x}{2} \right).
\end{eqnarray*}
So the sum is telescopic and we get $\color{red}{x=\cot(1)}$.
