The value of lim n → ∞ ∑ k = 1 n n 3 n 2 + k 2 n 2 + 3 k 2 is :

Given, $\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{n^3}{\left(n^2+k^2\right)\left(n^2+3k^2\right)}$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\frac{1}{\left(1+\frac{k^2}{n^2}\right)\left(1+3·\frac{k^2}{n^2}\right)}$

Now, using limit as a sum integral we get,

$={\int }_0^1\frac{dx}{\left(1+x^2\right)\left(1+3x^2\right)}$

$=\frac{1}{2}\left[{\int }_0^1\frac{3}{\left(1+3x^2\right)}-\frac{1}{\left(1+x^2\right)}\right]dx$

$=\frac{1}{2}\left[{\int }_0^1\frac{1}{\left({\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2+x^2\right)}-\frac{1}{\left(1+x^2\right)}\right]dx$

$=\frac{1}{2}{\left[\sqrt{3}{\tan }^{-1}\left(\sqrt{3}x\right)-{\tan }^{-1}x\right]}_0^1$

$=\frac{1}{2}\left[\sqrt{3}{\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}1\right]$

$=\frac{1}{2}\left[\sqrt{3}·\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\right]$

$=\frac{1}{2}\left[\frac{\mathrm{\pi }}{\sqrt{3}}-\frac{\mathrm{\pi }}{4}\right]$

$=\frac{13\pi }{8·(4\sqrt{3}+3)}$