<s>[INST] A problem on inverse trigonometric function How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{1+4n^2-1}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\\
&=\sum_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]\\
&=\tan^{-1}∞-\tan^{-1}1\\
&=\frac{π}{2}-\frac{π}{4}\\
&=\frac{π}{4}\end{align} Now one thing I have to confess that I knew this method previously,so I did and I don't understand from where this idea suddenly comes about telescoping? Can someone help me explain with any different approach is added as an answer. Thanks for your attention.
 [/INST] The series transformations might seem at first glance rather arbitrary. But there are some nice aspects behind  the curtain which can be sometimes conveniently used.
Let's assume we want to calculate
\begin{align*}
\color{blue}{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{2n^2}\right)}\tag{1}
\end{align*}
We know that telescoping is one of the techniques which can be sometimes used to tackle such problems. But is it plausible in this case?
Addition theorem:  Here we might recall the addition theorem for the tangent function and its inverse
\begin{align*}
\tan\left(\alpha\pm\beta\right)&=\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\tag{2.a}\\
\alpha\pm\beta &=\tan^{-1}\left(\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\right)\tag{2.b}
\end{align*}

Letting $x:=\tan \alpha$ and $y:=\tan\beta$ we obtain from (2.b) the identity
\begin{align*}
\color{blue}{\tan^{-1} x - \tan^{-1} y  = \tan^{-1}\left(\frac{x- y}{1+ xy}\right)}\tag{2.c}
\end{align*}

The point is, if we recall one of (2.a), (2.b) or (2.c) the telescoping approach becomes interesting, especially in it's most convenient representation (2.c).
Transformation: When looking at what we have in (1) the challenge is to find a transformation from
\begin{align*}
\tan^{-1}\frac{1}{2n^2}\quad\to\quad (2.c) \quad\to  \quad\tan^{-1} x - \tan^{-1} y
\end{align*}

We put the focus on the fraction $\frac{x-y}{1+xy}$ at (2.c) and obtain
\begin{align*}
\color{blue}{\frac{1}{2n^2}}&=\frac{1}{1+\left(2n^2-1\right)}\tag{3}\\
&=\frac{2}{1+\left(4n^2-1\right)}\tag{4}\\
&=\frac{2}{1+\left(2n+1\right)\left(2n-1\right)}\tag{5}\\
&\,\,\color{blue}{=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\tag{6}
\end{align*}

Comment:

*

*In (3) we write the denominator as $1+\left(2n^2-1\right)$ since we want as in (2.c)  something like $1+xy$. We see we  have $1+\left(a^2-b^2\right)=1+(a+b)(a-b)$. The factor $2$ in $\left(2n^2+1\right)$ is not that convenient, but we can cope with it by expanding by a factor $2$ to get $4n^2-1=(2n+1)(2n-1)$ which is done in (4).


*In (5) we have the denominator according to that in (2.c) and we are lucky, since the numerator $2$ can be written as $2=(2n+1)-(2n-1)$ to obtain in (6) the wanted representation as in (2.c).

Series calculation: Now we are well prepared to tackle the series by a telescoping approach. We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^\infty}\color{blue}{\tan^{-1}\frac{1}{2n^2}}
&=\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)\tag{$\to (6)$}\\
&=\sum_{n=1}^{\infty}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\tag{$\to (2.c)$}\\
&=\lim_{N\to\infty}\sum_{n=1}^{N}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\\
&=\lim_{N\to\infty}\left[\tan^{-1}\left(2N+1\right)-\tan^{-1}\left(1\right)\right]\\
&=\lim_{N\to\infty}\tan^{-1}\left(2N+1\right)-\lim_{N\to\infty}\tan^{-1}\left(1\right)\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\
&\,\,\color{blue}{=\frac{\pi}{4}}
\end{align*}

</s>