Are you ready for JEE-4?

$\begin{aligned}\mathfrak{T}_n=&\displaystyle\sum_{r=1}^n\tan^{-1}\left( \dfrac{1}{2r^2}\right)\\=&\displaystyle\sum_{r=1}^n\tan^{-1} \left(\dfrac{2}{ 4r^2}\right)\\=&\displaystyle\sum_{r=1}^n\tan^{-1}\left( \dfrac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)\\=&\displaystyle\sum_{r=1}^n \tan^{-1}(2r+1)-\tan^{-1}(2r-1)\end{aligned}$

$\large\mathbf{(A~Telescopic~Series)}$

$\large{ \mathfrak{T}_n=\tan^{-1}(2n+1)-\tan^{-1}(1)~~~~\\=\tan^{-1}(2n+1)-\dfrac{\pi}{4}}~~~$

$\large\therefore\displaystyle\lim_{n\to\infty}\mathfrak{T}_n=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$

$\huge \therefore \color{#D61F06}{1}+4(\color{#3D99F6}{4})=\boxed{17}$

$S = \displaystyle \lim_{n\rightarrow \infty} \sum_{k=1}^{n} \cot^{-1}(2k^{2})$

$S =\displaystyle \lim_{n\rightarrow \infty} \sum_{k=1}^{n} \tan^{-1}\left(\dfrac{1}{2k^{2}}\right) =\displaystyle \lim_{n\rightarrow \infty} \sum_{k=1}^{n} \tan^{1}\left(\dfrac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}\right)$
$\therefore S = \displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \tan^{-1}(2k+1) - \tan^{-1}(2k-1)$
$\therefore S = \displaystyle \lim_{n \rightarrow \infty} \tan^{-1}(3) - \tan^{-1}(1) + \tan^{-1}(5) - \tan^{-1}(3) + \ldots \tan^{-1}(2n+1) - \tan^{-1}(2n-1) = \displaystyle \lim_{n \rightarrow \infty} \tan^{-1}(2n+1) - \tan^{-1}(1) = \dfrac{\pi}{2} -\dfrac{\pi}{4} =\dfrac{\pi}{4}$
$\alpha + 4\gamma = 1 + 4\cdot 4 = 17$