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?
On
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.
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 $\color{red}{x=\cot(1)}$.