The value of _ n → ∞ ∑ _ k = 1 ^ n n ^ 3 ( n ^ 2 + k ^ 2 )( n ^ 2 +...

Given, $\lim_{n \to \infty}\mathop{∑}_{k = 1}^{n}\dfrac{{n}^{3}}{\left({n}^{2} + {k}^{2}\right)\left({n}^{2} + 3{k}^{2}\right)}$

$= \lim_{n \to \infty}\dfrac{1}{n}\mathop{∑}_{k = 1}^{n}\dfrac{1}{\left(1 + \dfrac{{k}^{2}}{{n}^{2}}\right)\left(1 + 3 \cdot \dfrac{{k}^{2}}{{n}^{2}}\right)}$

Now, using limit as a sum integral we get,

$= {∫}_{0}^{1}\dfrac{dx}{\left(1 + {x}^{2}\right)\left(1 + 3{x}^{2}\right)}$

$= \dfrac{1}{2}\left[{∫}_{0}^{1}\dfrac{3}{\left(1 + 3{x}^{2}\right)} - \dfrac{1}{\left(1 + {x}^{2}\right)}\right]dx$

$= \dfrac{1}{2}\left[{∫}_{0}^{1}\dfrac{1}{\left({\left(\dfrac{1}{\sqrt{3}}\right)}^{2} + {x}^{2}\right)} - \dfrac{1}{\left(1 + {x}^{2}\right)}\right]dx$

$= \dfrac{1}{2}{\left[\sqrt{3}\tan^{- 1}\left(\sqrt{3}x\right) - \tan^{- 1}x\right]}_{0}^{1}$

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

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

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

$= \dfrac{13\pi}{8 \cdot ( 4\sqrt{3} + 3 )}$