Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}& \sum _{\mathrm{r}=1}^{\infty }\frac{\mathrm{n}}{\sqrt{{\mathrm{n}}^4+{\mathrm{r}}^4}}-\frac{2{\mathrm{nr}}^2}{\left({\mathrm{n}}^2+{\mathrm{r}}^2\right)\sqrt{{\mathrm{n}}^4+{\mathrm{r}}^4}}\\ & \sum _{\mathrm{r}=1}^{\infty }\frac{\frac{1}{\mathrm{n}}}{\sqrt{1+{\left(\frac{\mathrm{r}}{\mathrm{n}}\right)}^4}}-\frac{2\left(\frac{1}{\mathrm{n}}\right){\left(\frac{\mathrm{r}}{\mathrm{n}}\right)}^2}{\left(1+{\left(\frac{\mathrm{r}}{\mathrm{n}}\right)}^2\right)\sqrt{1+{\left(\frac{\mathrm{r}}{\mathrm{n}}\right)}^4}}\\ & \Rightarrow {\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}-\frac{2{\mathrm{x}}^2\mathrm{dx}}{\left(1+{\mathrm{x}}^2\right)\sqrt{1+{\mathrm{x}}^4}}\\ & \Rightarrow {\int }_0^1\frac{1-{\mathrm{x}}^2}{\left(1+{\mathrm{x}}^2\right)\sqrt{1+{\mathrm{x}}^4}}dx\\ & \Rightarrow {\int }_0^1\frac{\frac{1}{{\mathrm{x}}^2}-1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\sqrt{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}}\mathrm{dx}\end{array}$ $\begin{array}{rl}& \Rightarrow -{\int }_0^1\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\sqrt{{\left(x+\frac{1}{x}\right)}^2-2}}dx\\ & x+\frac{1}{x}=t\Rightarrow 1-\frac{1}{x^2}dx=dt\end{array}$ $\begin{array}{rl}& \Rightarrow -{\int }_{\infty }^2\frac{\mathrm{dt}}{\mathrm{t}\sqrt{{\mathrm{t}}^2-2}}\\ & \Rightarrow -{\int }_{\infty }^2\frac{\mathrm{tdt}}{{\mathrm{t}}^2\sqrt{{\mathrm{t}}^2-2}}\\ & \text{ take }{\mathrm{t}}^2-2={\alpha }^2\\ & \mathrm{tdt}=\alpha \mathrm{d}\alpha \\ & \Rightarrow -{\int }_{\infty }^{\sqrt{2}}\frac{\alpha \mathrm{d}\alpha }{\left({\alpha }^2+2\right)\alpha }\\ & \Rightarrow -{\int }_{\infty }^{\sqrt{2}}\frac{\mathrm{d}\alpha }{{\alpha }^2+2}\end{array}$ $\begin{array}{rl}& {\Rightarrow \frac{-1}{\sqrt{2}}{\tan }^{-1}\frac{\alpha }{\sqrt{2}}]}_{\infty }^{\sqrt{2}}\\ & \Rightarrow \frac{-1}{\sqrt{2}}\left\{{\tan }^{-1}1\right\}+\frac{1}{\sqrt{2}}{\tan }^{-1}\infty \\ & \Rightarrow \frac{1}{\sqrt{2}}\left\{\frac{\pi }{2}-\frac{\pi }{4}\right\}\\ & \Rightarrow \frac{\pi }{4\sqrt{2}}=\frac{\pi }{\mathrm{K}}\end{array}$
So $\mathrm{K}=4\sqrt{2}$ ${\mathrm{K}}^2=32$