If ${\mathrm{U}}_{\mathrm{n}}=\left(1+\frac{1}{{\mathrm{n}}^2}\right){\left(1+\frac{2^2}{{\mathrm{n}}^2}\right)}^2\cdots {\left(1+\frac{{\mathrm{n}}^2}{{\mathrm{n}}^2}\right)}^{\mathrm{n}}$, then $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left({\mathrm{U}}_{\mathrm{n}}\right)}^{\frac{-4}{{\mathrm{n}}^2}}$ is equal to

$\begin{array}{l}\mathrm{Let} \mathrm{y}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left({\mathrm{U}}_{\mathrm{n}}\right)}^{\frac{-4}{{\mathrm{n}}^2}}\\ \mathrm{y}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left[{\left(1+\frac{1}{{\mathrm{n}}^2}\right)}^{\frac{-4}{{\mathrm{n}}^2}}{\left(1+\frac{2^2}{{\mathrm{n}}^2}\right)}^{\frac{-4}{{\mathrm{n}}^2}\cdot 2}{\left(1+\frac{3^2}{{\mathrm{n}}^2}\right)}^{\frac{-4}{{\mathrm{n}}^2}\cdot 3}\cdots \right]\end{array}$

Taking log on both sides, we get

          $$\begin{gathered} \ln \mathrm{y}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{r}=1}^{\mathrm{n}} \left[\frac{-4}{{\mathrm{n}}^2}\cdot \mathrm{rln}\left(1+\frac{{\mathrm{r}}^2}{{\mathrm{n}}^2}\right)\right] \\ \Rightarrow \ln \mathrm{y}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \underset{\mathrm{r}=1}{\overset{\mathrm{n}}{\sum \left(-4\right)}} \left[\frac{1}{\mathrm{n}}\cdot \frac{r}{n} \ln \left(1+\frac{{\mathrm{r}}^2}{{\mathrm{n}}^2}\right)\right] \end{gathered}$$

Now, replace $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \mathrm{\Sigma }\to \int$

        $\frac{\mathrm{r}}{\mathrm{n}}\to \mathrm{x}, \frac{1}{\mathrm{n}}\to \mathrm{dx}$

Lower limit = 0

Upper limit = 1

$\begin{array}{l}\therefore \ln \mathrm{y}={\int }_0^1 -4\mathrm{xln}\left(1+{\mathrm{x}}^2\right)\mathrm{dx}\\ \text{ Let }1+{\mathrm{x}}^2=\mathrm{t}\\ \Rightarrow \mathrm{xdx}=\frac{\mathrm{dt}}{2}\\ \text{ When }\mathrm{x}\to 0, \mathrm{t}\to 1\\ \text{ and }\mathrm{x}\to 1, \mathrm{t}\to 2\\ \therefore \ln \mathrm{y}={\int }_1^2 -2\ln \mathrm{tdt}\\ =-2(\mathrm{tln}\mathrm{t}-\mathrm{t}{)}_1^2\\ =-2(2\ln 2-2+1)\\ =-2(2\ln 2-1)\\ \Rightarrow \ln \mathrm{y}=\ln \frac{1}{16}+2\\ \Rightarrow \mathrm{y}=\frac{1}{16}{\mathrm{e}}^2\end{array}$