Step 1: Convert $\cot^{-1}$ to $\tan^{-1}$.
Recall that $\cot^{-1}(x) = \tan^{-1}\!\left(\dfrac{1}{x}\right)$ for $x > 0$. The general term becomes: \[\cot^{-1}\!\left(r^2 + \tfrac{3}{4}\right) = \tan^{-1}\!\left(\frac{4}{4r^2+3}\right)\]
Step 2: Write the numerator as a difference and use the $\tan^{-1}$ subtraction formula.
Notice that $4r^2 + 3 = (r^2 - 1/4) + (9/4) = (r-1/2)(r+1/2) + 1 + 0$... more directly: we claim: \[\tan^{-1}\!\left(r+\tfrac{1}{2}\right) - \tan^{-1}\!\left(r-\tfrac{1}{2}\right) = \tan^{-1}\!\left(\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}\right)\] The numerator inside is $1$, and $1 + (r^2 - 1/4) = r^2 + 3/4$. So: \[\tan^{-1}\!\left(r+\tfrac{1}{2}\right) - \tan^{-1}\!\left(r-\tfrac{1}{2}\right) = \tan^{-1}\!\left(\frac{1}{r^2+3/4}\right) = \cot^{-1}\!\left(r^2+\tfrac{3}{4}\right)\]
Step 3: Write the partial sum using telescoping.
\[S_n = \sum_{r=1}^{n}\left[\tan^{-1}\!\left(r+\tfrac{1}{2}\right) - \tan^{-1}\!\left(r-\tfrac{1}{2}\right)\right]\] This is a telescoping sum. All intermediate terms cancel: \[S_n = \tan^{-1}\!\left(n+\tfrac{1}{2}\right) - \tan^{-1}\!\left(\tfrac{1}{2}\right)\]
Step 4: Take the limit as $n \to \infty$.
\[\lim_{n\to\infty} \tan^{-1}\!\left(n+\tfrac{1}{2}\right) = \frac{\pi}{2}\] So: \[S = \frac{\pi}{2} - \tan^{-1}\!\left(\frac{1}{2}\right)\]
Step 5: Simplify using the complementary angle identity.
For $x > 0$: $\dfrac{\pi}{2} - \tan^{-1}(x) = \tan^{-1}\!\left(\dfrac{1}{x}\right)$. With $x = \dfrac{1}{2}$: \[S = \tan^{-1}\!\left(\frac{1}{1/2}\right) = \tan^{-1}(2)\]
Step 6: Confirm the telescoping is correct.
When $r=1$: $\tan^{-1}(3/2) - \tan^{-1}(1/2)$. When $r=2$: $\tan^{-1}(5/2) - \tan^{-1}(3/2)$. The $\tan^{-1}(3/2)$ terms cancel. This continues, leaving only $\tan^{-1}(n+1/2) - \tan^{-1}(1/2)$ as the partial sum.
Step 7: State the final answer.
\[ \boxed{\tan^{-1} 2} \]