If U n = 1 + 1 n 2 1 + 2 2 n 2 2 … 1 + n 2 n 2 n , then lim n → ∞ U n - 4 n 2 is equal to

$U_n=\prod _{r=1}^n{\left(1+\frac{r^2}{n^2}\right)}^r$

$L=\underset{n\to \infty }{\lim }{\left(U_n\right)}^{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$n^2$}\right.}$

$\log L=\underset{n\to \infty }{\lim }\frac{-4}{n^2}\sum _{r=1}^n \log {\left(1+\frac{r^2}{n^2}\right)}^r$

$\log L=\underset{n\to \infty }{\lim } \sum _{r=1}^n \frac{-4r}{n}·\frac{1}{n} \log \left(1+\frac{r^2}{n^2}\right)$

$\log L=-4{\int }_0^{1}x \log \left(1+x^2\right)dx$

Put $1+x^2=t$

$2x dx=dt$

$=-2{\int }_1^2\log \left(t\right)dt=-2{\left[t \log t-t\right]}_1^2$

$\Rightarrow \log L=-2\left(2\log 2-1\right)$

$\therefore L=e^{-2\left(2\log 2-1\right)}$

$=e^{-2\left(\log \frac{4}{e}\right)}=e^{\log {\left(\frac{4}{e}\right)}^{-2}}$

$L= {\left(\frac{e}{4}\right)}^2=\frac{e^2}{16}$