Question

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

• 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 $l=\underset{n\to \infty }{\lim }{\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)+\dots \dots +\left(1+\frac{n^2}{n^2}\right)\right]}^{1/n}$
$logl=\underset{n\to \infty }{\overset{l/n}{\lim }}\left[log\left(1+\frac{1}{n^2}\right)+log\left(1+\frac{2^2}{n^2}\right)+\dots +log\left(1+\frac{n^2}{n^2}\right)\right]$
$\therefore =\underset{h\to 0}{\lim }\sum h\cdot log{\left(1+rh\right)}^2)$
$=\underset{0}{\overset{1}{\int }}log\left(1+x^2\right)dx$
$={\left|log\left(1+x^2\right)\cdot x\right|}_0^1-\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}\cdot 2x\cdot xdx$
$=log2-2\underset{0}{\overset{1}{\int }}\frac{x^2}{1+x^2}dx$
$=log2-2\underset{0}{\overset{1}{\int }}\left(1-\frac{1}{1+x^2}\right)dx$
$=log2-2{\left[xta{n}^{-1}x\right]}_0^1$
$=log2-2\left[1-\frac{\pi }{4}\right]=log2+\frac{\pi -4}{2}$
$\Rightarrow log\frac{l}{2}=\frac{\pi -4}{2}$
$\therefore l=2e\frac{\pi -4}{2}$