The value of n lim _k=1^nn^3(n^2+k^2)(n^2+3k^2) is :

Question

The value of $\underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{{n}^{3}}{({n}^{2}+{k}^{2})({n}^{2}+3{k}^{2})}$ is :

Accepted Answer

$\begin{gathered} \underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{{n}^{3}}{{n}^{4}(1+\frac{{k}^{2}}{{n}^{2}})(1+\frac{3{k}^{2}}{{n}^{2}})} \ =\underset{n	o \infty}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{{n}^{3}}{(1+\frac{{k}^{2}}{{n}^{2}})(1+\frac{3{k}^{2}}{{n}^{2}})} \end{gathered}$

$=\int _{0}^{1}\frac{dx}{3(1+{x}^{2})(\frac{1}{3}+{x}^{2})}$

$\begin{gathered} =\int _{0}^{1}\frac{1}{3}	imes \frac{3}{2}\frac{({x}^{2}+1)-({x}^{2}+\frac{1}{3})}{(1+{x}^{2})({x}^{2}+\frac{1}{3})}dx \ =\frac{1}{2}\int _{0}^{1}[[|\frac{1}{{x}^{2}+{(\frac{1}{\sqrt{3}})}^{2}}-\frac{1}{1+{x}^{2}}]]|dx \ =\frac{1}{2}{[\sqrt{3}{	an }^{-1}(\sqrt{3}x)]}_{1}^{0}-\frac{1}{2}{({	an }^{-1}x)}_{1}^{0} \ =\frac{\sqrt{3}}{2}(\frac{\pi }{3})-\frac{1}{2}(\frac{\pi }{4})=\frac{\pi }{2\sqrt{3}}-\frac{\pi }{8} \ =\frac{13\pi }{8\cdot (4\sqrt{3}+3)} \end{gathered}$

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