Question

$\displaystyle\lim_{n\to \infty} \left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)....\left(1+\frac{n^n}{n^2}\right)\right]^{1/n}=$

• A. $e^{\frac{\pi-4}{2}}$
• B. $2e^{\frac{\pi-4}{2}}$
• C. $\frac{e^{\frac{\pi-4}{2}}}{2}$
• D. none of these.

Answer 1

Answer: (B)

Let $l= \displaystyle\lim_{n\to\infty} \left[\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{2^{2}}{n^{2}}\right)+\ldots\ldots+\left(1+\frac{n^{2}}{n^{2}}\right)\right]^{1 /n}$
$log\,l= \displaystyle\lim^{l /n}_{n\to\infty} \left[log \left(1+\frac{1}{n^{2}}\right)+log \left(1+\frac{2^{2}}{n^{2}}\right)+\ldots+log\left(1+\frac{n^{2}}{n^{2}}\right)\right]$
$\therefore =\displaystyle \lim_{h\to0} \sum h\cdot log \left(1+rh\right)^{2})$
$=\int\limits_{0}^{1}log \left(1+x^{2}\right)dx$
$=\left|log \left(1+x^{2}\right)\cdot x\right|_{0}^{1}-\int\limits_{0}^{1} \frac{1}{1+x^{2}}\cdot2x\cdot x dx$
$=log\,2-2\int\limits_{0}^{1}\frac{x^{2}}{1+x^{2}}dx$
$=log\,2-2 \int\limits_{0}^{1}\left(1-\frac{1}{1+x^{2}}\right)dx$
$=log\,2-2\left[x\,tan^{-1}x\right]_{0}^{1}$
$=log\,2-2\left[1-\frac{\pi}{4}\right]=log\,2+\frac{\pi-4}{2}$
$\Rightarrow log\frac{l}{2} =\frac{\pi-4}{2}$
$\therefore l=2e \frac{\pi-4}{2}$