$\begin{aligned} & t_{n}=\cot ^{-1} 2 n^{2} \\=& \tan ^{-1} \frac{1}{2 n^{2}}=\tan ^{-1} \frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)} \\=& \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1) \\ \therefore & S_{n}=\tan ^{-1}(2 n+1)-\tan ^{-1} 1 \\ \therefore & \operatorname{Lim}_{n \rightarrow \infty} S_{n}=\pi / 2-\pi / 4=\pi / 4 \end{aligned}$