I was trying to solve up this equation but couldn't move ahead. $$\sum_ {n=1}^{\infty} \cot^{-1}(2n^2)$$ I wrote the expression as $$\sum_ {n=1}^{\infty} \tan^{-1}\left( \frac{1}{2n^2}\right)$$
I wanted to change the expression into such a form such that it can take up the form of $\tan^{-1}A-\tan^{-1}B$ so that all the terms except the second one get cancelled up but I am unable to think of any manipulation through which I can get the thing done.
Can anybody give me a hint on how to go ahead?
$$\sum_ {n=1}^{\infty} \tan^{-1}\left( \frac{1}{2n^2}\right)=\sum_ {n=1}^{\infty} \tan^{-1}\left( \frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}\right)$$ $$\sum_ {n=1}^{\infty} \tan^{-1}\left( \frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}\right)=\sum_ {n=1}^{\infty}\tan^{-1}(2n+1)-\tan^{-1}(2n-1)=\frac{\pi}{4}$$
To recognize a telescopic sum as Behrouz did is the key for a simple proof.
I will go for the overkill. We have
$$ \sum_{n\geq 1}\arctan\frac{1}{2n^2}\leq \sum_{n\geq 1}\frac{1}{2n^2}=\frac{\pi^2}{12}<1\tag{1}$$
hence:
$$ \sum_{n\geq 1}\arctan\frac{1}{2n^2}=\text{Arg}\prod_{n\geq 1}\left(1+\frac{i}{2n^2}\right) \tag{2} $$
and since the Weierstrass product for the $\sinh$ function gives
$$ \prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)=\frac{\sinh(z\pi)}{z\pi}\tag{3} $$
we have:
$$\sum_{n\geq 1}\arctan\frac{1}{2n^2}=\text{Arg}\left(\frac{\cosh\frac{\pi}{2}}{\pi}(1+i)\right)=\text{Arg}(1+i)=\color{red}{\frac{\pi}{4}}.\tag{4}$$
The advantage of this approach is that it computes $$ \sum_{n\geq 1}\arctan\frac{1}{n^2}=\frac{\pi}{4}-\arctan\left(\frac{\tanh\frac{\pi}{\sqrt{2}}}{\tan\frac{\pi}{\sqrt{2}}}\right)\tag{5} $$ (and similar ones) too, where it is not easy at all to write the main of the LHS as a telescopic term.
With $$2n^2=\frac{\Big(\frac{1}{n-1}+1\Big)\Big(\frac{1}{n}+1\Big)+1}{\Big(\frac{1}{n-1}+1\Big)-\Big(\frac{1}{n}+1\Big)}$$ we obtain $$\cot^{-1}(2n^2)=\cot^{-1}\Big(\frac{1}{n}+1\Big)-\cot^{-1}\Big(\frac{1}{n-1}+1\Big)$$ Hence $$\sum_1^N\cot^{-1}(2n^2)=\cot^{-1}\Big(\frac{1}{N}+1\Big).$$
