If U_{n}=left(1+frac{1^{2}}{n^{2}}right)^{1}l | vedclass

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(A) Given $U_{n}=\prod_{r=1}^{n}\left(1+\frac{r^{2}}{n^{2}}\right)^{r}$.Let $L=\lim _{n \rightarrow \infty}\left(U_{n}\right)^{-4 / n^{2}}$.Taking nat

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$\frac{e^{2}}{16}$

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(A) Given $U_{n}=\prod_{r=1}^{n}\left(1+\frac{r^{2}}{n^{2}}\right)^{r}$.Let $L=\lim _{n \rightarrow \infty}\left(U_{n}\right)^{-4 / n^{2}}$.Taking natural logarithm on both sides:$\log L=\lim _{n \rightarrow \infty} \frac{-4}{n^{2}} \sum_{r=1}^{n} r \log \left(1+\frac{r^{2}}{n^{2}}\right)$.We can rewrite this as:$\log L=\lim _{n \rightarrow \infty} \sum_{r=1}^{n}-4 \left(\frac{r}{n}\right) \log \left(1+\left(\frac{r}{n}\right)^{2}\right) \cdot \frac{1}{n}$.This is a Riemann sum,which converts to the definite integral:$\log L = -4 \int_{0}^{1} x \log (1+x^{2}) \, dx$.Let $t = 1+x^{2}$,then $dt = 2x \, dx$.When $x=0, t=1$; when $x=1, t=2$.$\log L = -2 \int_{1}^{2} \log (t) \, dt = -2 [t \log t - t]_{1}^{2}$.$\log L = -2 [(2 \log 2 - 2) - (1 \log 1 - 1)] = -2 [2 \log 2 - 2 + 1] = -2 [2 \log 2 - 1]$.$\log L = -4 \log 2 + 2 = \log (2^{-4}) + \log (e^{2}) = \log \left(\frac{e^{2}}{16}\right)$.Therefore,$L = \frac{e^{2}}{16}$.

If $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$,then $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ is equal to :

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