The value of $$\lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}$$ is :

<p>$$\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \end{aligned}$$</p> <p>$=\int_\limits0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}$</p> <p>$$\begin{aligned} & =\int_\limits0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\ & =\frac{1}{2} \int_\limits0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\ & =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\ & =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\ & =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)} \end{aligned}$$</p>