Solving an Inverse-Trigonometric Limit!

$L = \lim_{n \to \infty} \left[ \dfrac{1}{n^2} \sum_{1 \leq i < j \leq n} \tan^{-1} \left ( \dfrac{i}{n} \right) \tan^{-1} \left ( \dfrac{j}{n} \right) \right]$

$\Rightarrow \lim_{n \to \infty} \dfrac{1}{2n^2} \left[ \left( \sum_{i=1}^n \tan^{-1}\left(\dfrac{i}{n}\right)\right)^2 - \left( \sum_{i=1}^n \left(\tan^{-1}\left(\dfrac{i}{n}\right)\right)^2\right) \right]$

$= \dfrac12 \left( \int_0^1 \tan^{-1}(x) \ dx \right)^2 - \dfrac{1}{2n} \int_0^1 \left(\left(\tan^{-1}(x)\right)^2\right)\ dx$

Since $n \to \infty$ , we obtain:

$L= \dfrac12 \left( \int_0^1 \tan^{-1}(x) \ dx \right)^2 = \dfrac12 \left(\dfrac{\pi - \ln(4)}{4} \right)^2$

So $A=1, B=4, C=2, D=5, A+B+C+D = \boxed{12}$ .