Evaluate $\sum_ {n=1}^{\infty} \cot^{-1}(2n^2)$

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.