Question

If $U_n=(1+\frac{1}{n^2}){(1+\frac{2^2}{n^2})}^2\dots {(1+\frac{n^2}{n^2})}^n$, then $\underset{n\to \infty }{lim}(U_n{)}^{\frac{-4}{n^2}}$ is equal to

• A. $\frac{4}{e^2}$
• B. $\frac{4}{e}$
• C. $\frac{16}{e^2}$
• D. $\frac{e^2}{16}$

Answer 1

The correct option is D $\frac{e^2}{16}$

Given:$U_n=(1+\frac{1}{n^2}){(1+\frac{2^2}{n^2})}^2\dots {(1+\frac{n^2}{n^2})}^n$

Let $L=\underset{n\to \infty }{lim}(U_n{)}^{\frac{-4}{n^2}}$
Taking $\ln$ on both sides, we get
$\ln L=\underset{n\to \infty }{lim}-\frac{4}{n^2}(\ln (1+\frac{1}{n^2})+2\ln (1+\frac{2^2}{n^2})+3\ln (1+\frac{3^2}{n^2})+\dots +n\ln (1+\frac{n^2}{n^2}))$
$\Rightarrow \ln L=-4\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{r}{n}\ln (1+{\left(\frac{r}{n}\right)}^2)$
$\Rightarrow \ln L=-4\underset{0}{\overset{1}{\int }}x\ln (1+x^2)dx$
$\Rightarrow \ln L=-2\underset{0}{\overset{1}{\int }}2x\ln (1+x^2)dx$
Put $1+x^2=t\Rightarrow 2xdx=dt$
$\Rightarrow \ln L=-2\underset{1}{\overset{2}{\int }}\left(\ln t\right)dt$
$\Rightarrow \ln L=-2[t\ln t-t{]}_1^2$
$\Rightarrow \ln L=-2(2\ln 2-1)$
$\Rightarrow \ln L=2(\ln e-\ln 4)=\ln \left(\frac{e^2}{16}\right)$
$\therefore L=\frac{e^2}{16}$