Telescopical?

Important identities: $\cot^{-1} x = \tan^{-1} \frac {1}{x}$ ; $\tan ^{-1} ∞ =\frac {π}{2}$ ; $\tan^{-1} x + \cot^{-1} x =\frac {π}{2}$

The series is $=\tan^{-1} \frac {12}{36r^2-5}$

Dividing by 4,

$=\tan ^{-1} \frac {3}{1+(9r^2-\frac {9}{4})}$

$=\tan ^{-1} \frac {(3r+1.5)-(3r-1.5)}{1+(3r+1.5)(3r-1.5)}=\tan ^{-1} (3r+1.5)- \tan ^{-1} (3r-1.5)$

This would form a telescopic series, where each term will be cancelled out by the following terms, leaving only $\tan ^{-1} ∞ - \tan ^{-1} (1.5) = \cot ^{-1} 1.5$

Relevant wiki: Writing a Proof by Induction

If we defined $\displaystyle S_n = \sum_{r=1}^n \cot^{-1} \left(3r^2-\frac 5{12}\right)$ , where $n$ is a positive integer, then we find that $S_n = \cot^{-1} \left(\dfrac {13}{12n} + \dfrac 32\right)$ . Let us prove by induction that the claim is true for all $n \ge 1$ .

For $n=1$ , $S_1 = \cot^{-1} \left(3(1^2) - \dfrac 5{12}\right) = \cot^{-1} \left(\dfrac {31}{12}\right) = \cot^{-1} \left(\dfrac {13}{12(1)} + \dfrac 32 \right)$ . The claim is true for $n=1$ .

Assuming the claim is true for $n$ , then

$\begin{aligned} S_{n+1} & = S_n + \cot^{-1} \left(3(n+1)^2 - \frac 5{12}\right) \\ & = \cot^{-1} \left(\frac {13}{12n} + \frac 32\right) + \cot^{-1} \left(3(n+1)^2 - \frac 5{12}\right) \\ & = \cot^{-1} \left(\frac {18n + 13}{12n} \right) + \cot^{-1} \left(\frac {36(n+1)^2 - 5}{12}\right) \\ & = \cot^{-1} \left(\frac {18(n +1)-5}{12n} \right) + \cot^{-1} \left(\frac {36(n+1)^2-5}{12}\right) \\ & = \cot^{-1} \left(\frac {\frac {18(n +1)-5}{12n} \times \frac {36(n+1)^2-5}{12} -1}{\frac {18(n +1)-5}{12n} + \frac {36(n+1)^2-5}{12}} \right) \\ & = \cot^{-1} \left(\frac {648(n+1)^3-180(n+1)^2-234(n+1)+169}{432(n+1)^3-432(n+1)^2+156(n+1)^2}\right) \\ & = \cot^{-1} \left(\frac 32 + \frac {13}{12(n+1)}\right) \end{aligned}$

Showing that the claim is also true for $n+1$ , therefore, it is true for all $n \ge 1$ .

Now we have $\displaystyle \lim_{n \to \infty} S_n = \lim_{n \to \infty} \cot^{-1} \left(\frac {13}{12n} + \frac 32\right) = \cot^{-1} \frac 32$ . $\implies 4A = 4 \times \dfrac 32 = \boxed 6$ .