Inspired by my Best Friend

Let $S=\cot^{-1} 2+\cot^{-1} 8+\cot^{-1} 18+\cdots$ , then $S=\displaystyle \sum_{n=1}^{\infty} \cot^{-1} (2n^2)$ or $S=\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\dfrac{1}{2n^2}\right)$ .

Then notice that $\dfrac{1}{2n^2}=\dfrac{2}{4n^2}=\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$ . With this we can make this sum telescope:

$S=\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)$

Finally, we apply the identity $\tan^{-1}(a)-\tan^{-1}(b)=\tan^{-1}\left(\dfrac{a-b}{1+ab}\right)$

$S=\displaystyle \sum_{n=1}^{\infty}(\tan^{-1}(2n+1)-\tan^{-1}(2n-1))$

Then, we are only left with this two terms:

$S=\tan^{-1}(\infty)-\tan^{-1}(1)=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\boxed{\dfrac{\pi}{4}}$