<s>[INST] 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?
 [/INST] Arguing loosly,
$\begin{array}\\
S_n
&=\sum_{i=1}^{n} arccot(\frac{(i+1)(i+2)x}{2}+\frac{2}{x})\\
&\approx \frac{2}{x}\sum_{i=1}^{n} \dfrac1{(i+1)(i+2)}
\qquad\text{since }arccot(x) \approx \frac1{x} \text{ for large } x \\
&=\frac{2}{x}\sum_{i=1}^{n} (\dfrac1{i+1}-\dfrac1{i+2})\\
&=\frac{2}{x}(\frac12-\frac1{n+2})
\\
&=\frac1{x}-\frac{2}{x(n+2)})
\\
\end{array}
$
So if $S_n \to 1$,
must have $x = 1$.
It would be entertaining 
if this sloppy argument
was correct.
</s>