If $U_{n}=\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \cdots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$, then $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ is equal to :
Correct Option: 1
$\mathrm{U}_{\mathrm{n}}=\prod_{\mathrm{r}=1}^{\mathrm{n}}\left(1+\frac{\mathrm{r}^{2}}{\mathrm{n}^{2}}\right)^{\mathrm{r}}$
$L=\lim _{n \rightarrow \infty}\left(U_{n}\right)^{-4 / n^{2}}$
$\log L=\lim _{n \rightarrow \infty} \frac{-4}{n^{2}} \sum_{r=1}^{n} \log \left(1+\frac{r^{2}}{n^{2}}\right)^{r}$
$\Rightarrow \log L=\lim _{n \rightarrow \infty} \sum_{r=1}^{n}-\frac{4 r}{n} \cdot \frac{1}{n} \log \left(1+\frac{r^{2}}{n^{2}}\right)$
$\Rightarrow \log \mathrm{L} \Rightarrow-4 \int_{0}^{1} \mathrm{x} \log \left(1+\mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}$
put $1+x^{2}=t$
Now, $2 x d x=d t$
$=-2 \int_{1}^{2} \log (t) d t=-2[t \log t-t]_{1}^{2}$
$\Rightarrow \log \mathrm{L}=-2(2 \log 2-1)$
$\therefore \mathrm{L}=\mathrm{e}^{-2(2 \log 2-1)}$
$=e^{-2\left(\log \left(\frac{4}{e}\right)\right)}$
$=e^{\log \left(\frac{4}{e}\right)^{-2}}$
$=\left(\frac{e}{4}\right)^{2}=\frac{e^{2}}{16}$
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