Question

If $f\left(n\right)=\frac{1}{n}\left\{\right(2n+1\left)\right(2n+2)\cdots (2n+n){\}}^{1/n}$, then $\underset{n\to \infty }{lim}f\left(n\right)$ equals

• A. 4/e
• B. 27/4e
• C. 27e/4
• D. 4e

Answer 1

The correct option is B 27/4e

$\text{Let}A=\frac{1}{n}\left\{\right(2n+1\left)\right(2n+2)\dots \dots -(2n+n){\}}^{1/n}$

$\log A=\frac{1}{n}\underset{n\to \infty }{lim}\log [\frac{(2n+1)}{n}\frac{(2n+2)}{n}\frac{(2n+3)}{n}\dots \dots \dots \dots ..\frac{(2n+n)}{n}]$
$\log A=\frac{1}{n}\underset{n\to \infty }{lim}\log [(2+\frac{1}{n})(2+\frac{2}{n})(2+\frac{3}{n})\dots \dots \dots \dots \dots \dots \dots (2+\frac{n}{n})]$
$\log A={\int }_0^1\log (2+x)dx$
$\log A=|x\log (2+x)-x+2\log (2+x){|}_0^1$
$\log A=[\log 3-1+2\log 3-2\log 2]$
$\log A=[\log 3-\mathrm{loge}+\log 9-\log 4]$
$\log A=\left[\log \right(27/4e\left)\right]$
$A=\left(\frac{27}{4e}\right)$