The value of $\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{\left(1^1+2^2+\dots +{\mathrm{n}}^2\right)\left(1^3+2^3+\dots +{\mathrm{n}}^3\right)\left(1^4+2^4+\dots +{\mathrm{n}}^4\right)}{{\left(1^5+2^5+\dots .+{\mathrm{n}}^5\right)}^2}$ is equal to

$$\begin{gathered} \underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{\left(1^1+2^2+\dots +{\mathrm{n}}^2\right)\left(1^3+2^3+\dots +{\mathrm{n}}^3\right)\left(1^4+2^4+\dots +{\mathrm{n}}^4\right)}{{\left(1^5+2^5+\dots .+{\mathrm{n}}^5\right)}^2} \\ \underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{n^2\left({\left({\displaystyle \frac{1}{n}}\right)}^1+{\left(\frac{2}{n}\right)}^2+\dots +{\left(\frac{\mathrm{n}}{\mathrm{n}}\right)}^2\right)n^3\left({\left(\frac{1}{n}\right)}^3+{\left(\frac{2}{n}\right)}^3+\dots +{\left(\frac{\mathrm{n}}{\mathrm{n}}\right)}^3\right)n^4\left({\left(\frac{1}{n}\right)}^4+{\left(\frac{2}{n}\right)}^4+\dots +{\left(\frac{\mathrm{n}}{\mathrm{n}}\right)}^4\right)}{n^{10}{\left({\left(\frac{1}{n}\right)}^5+{\left(\frac{2}{n}\right)}^5+\dots .+{\left(\frac{\mathrm{n}}{\mathrm{n}}\right)}^5\right)}^2} \\ \underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{{\displaystyle \frac{1}{n}}\sum _{r=1}^{r=n}{\left({\displaystyle \frac{r}{n}}\right)}^2.{\displaystyle \frac{1}{n}}\sum _{r=1}^{r=n}{\left(\frac{r}{n}\right)}^3.{\displaystyle \frac{1}{n}}\sum _{r=1}^{r=n}{\left(\frac{r}{n}\right)}^4}{{\left[ {\displaystyle \frac{1}{n}}\sum _{r=1}^{r=n}{\left(\frac{r}{n}\right)}^5\right]}^2} \end{gathered}$$

$=\frac{{\int }_0^1 {\mathrm{x}}^2\mathrm{dx}\cdot {\int }_0^1 {\mathrm{x}}^3\mathrm{dx}\cdot {\int }_0^1 {\mathrm{x}}^4\mathrm{dx}}{{\left({\int }_0^1 {\mathrm{x}}^5\mathrm{dx}\right)}^2}=\frac{\frac{1}{3}\times \frac{1}{4}\times \frac{1}{5}}{\frac{1}{6}\times \frac{1}{6}}=\frac{36}{60}=\frac{3}{5}$