The value of $$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)}$$ is :

We evaluate $$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)}$$.

Let $$t = \frac{k}{n}$$ so that the sum becomes a Riemann sum given by $$\sum_{k=1}^n \frac{n^3}{n^2(1+t^2)\,n^2(1+3t^2)}\cdot 1 = \sum_{k=1}^n \frac{1}{n}\,\frac{1}{(1+t^2)(1+3t^2)}\,,\quad t=\frac{k}{n}\,,$$ and as $$n\to\infty$$ this tends to $$I = \int_0^1 \frac{dt}{(1+t^2)(1+3t^2)}\,. $$

Using partial fractions one writes $$\frac{1}{(1+t^2)(1+3t^2)} = \frac{A}{1+t^2} + \frac{B}{1+3t^2}\,, $$ and expanding gives $$1 = A(1+3t^2)+B(1+t^2)\,. $$ Equating constant terms and coefficients of $$t^2$$ yields $$A+B=1\,,\quad 3A+B=0\,, $$ so that $$A=-\tfrac12\,,\ B=\tfrac32\,. $$

Hence $$I = \int_0^1\Bigl[\frac{-\tfrac12}{1+t^2}+\frac{\tfrac32}{1+3t^2}\Bigr]dt = -\frac12\int_0^1\frac{dt}{1+t^2} + \frac32\int_0^1\frac{dt}{1+3t^2}\,, $$ and integrating gives $$-\frac12\bigl[\tan^{-1}(t)\bigr]_0^1 + \frac32\cdot\frac{1}{\sqrt3}\bigl[\tan^{-1}(\sqrt3\,t)\bigr]_0^1 = -\frac12\cdot\frac\pi4 + \frac{\sqrt3}{2}\cdot\frac\pi3 = -\frac{\pi}{8} + \frac{\sqrt3\pi}{6} = \frac{\pi(4\sqrt3-3)}{24}\,. $$

We compare this with option (2), namely $$\frac{13\pi}{8(4\sqrt3+3)}\,. $$ Rationalizing the denominator shows $$\frac{13\pi}{8(4\sqrt3+3)}\cdot\frac{4\sqrt3-3}{4\sqrt3-3} = \frac{13\pi(4\sqrt3-3)}{8(48-9)} = \frac{\pi(4\sqrt3-3)}{24}\,,$$ so the two expressions agree.

The answer is Option (2): $$\boxed{\frac{13\pi}{8(4\sqrt3+3)}}$$.