$\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \infty =$

The infinite series S is defined as $S = \sum_{n=1}^{\infty} \cot^{-1}(2n^2)$.

For positive arguments, $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$, so the general term becomes $T_n = \tan^{-1}\left(\frac{1}{2n^2}\right)$.

Multiplying numerator and denominator by 2 yields $T_n = \tan^{-1}\left(\frac{2}{4n^2}\right)$.

Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A - B}{1 + AB}\right)$ with $A = 2n + 1$ and $B = 2n - 1$ gives $T_n = \tan^{-1}(2n + 1) - \tan^{-1}(2n - 1)$, since $A - B = 2$ and $1 + AB = 4n^2$.

The partial sum telescopes when expanded:

$S_k = \sum_{n=1}^{k} \left[\tan^{-1}(2n + 1) - \tan^{-1}(2n - 1)\right] = \tan^{-1}(2k + 1) - \tan^{-1}(1)$.

Taking the limit as $k \to \infty$, $\tan^{-1}(2k + 1) \to \frac{\pi}{2}$, and since $\tan^{-1}(1) = \frac{\pi}{4}$, the infinite sum evaluates to $S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

Answer: $\frac{\pi}{4}$