Step 1: Understanding the Question:
This is an infinite series involving inverse trigonometric functions. We use a telescoping series method to solve it. Step 3: Detailed Explanation:
1. General term $T_n = \cot^{-1}(2n^2) = \tan^{-1}(\frac{1}{2n^2})$.
2. Use the identity $\tan^{-1}(\frac{1}{2n^2}) = \tan^{-1}(\frac{2}{4n^2}) = \tan^{-1}(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)})$.
3. This is in the form $\tan^{-1}(\frac{x-y}{1+xy}) = \tan^{-1} x - \tan^{-1} y$.
So, $T_n = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
4. Sum of first $n$ terms $S_n$:
$S_n = [\tan^{-1} 3 - \tan^{-1} 1] + [\tan^{-1} 5 - \tan^{-1} 3] + \dots + [\tan^{-1}(2n+1) - \tan^{-1}(2n-1)]$
All terms cancel except the last positive and first negative:
$S_n = \tan^{-1}(2n+1) - \tan^{-1} 1$.
5. For infinite series, take $n \rightarrow \infty$:
$S_\infty = \tan^{-1}(\infty) - \tan^{-1} 1 = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. Step 4: Final Answer:
The sum of the series is $\frac{\pi}{4}$.