What is $x$, if $\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots = 1$?

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

Tags:

Trigonometry

Calculus

Sequences And Series

Telescopic Series

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