Question

The value of $\underset{n\to \infty }{lim}\frac{\left[\right(n+1\left)\right(n+2)...(n+n){]}^{\frac{1}{n}}}{n}$ is:

• A. $\frac{4}{e^2}$
• B. $\frac{3}{e}$
• C. $\frac{3}{e^2}$
• D. $\frac{4}{e}$

Answer 1

The correct option is D $\frac{4}{e}$

Let $L=\underset{n\to \infty }{lim}\frac{\left[\right(n+1\left)\right(n+2)\cdots (n+n){]}^{1/n}}{n}$
$=\underset{n\to \infty }{lim}{\left[\frac{(n+1)(n+2)\cdots (n+n)}{n^n}\right]}^{1/n}$
$=\underset{n\to \infty }{lim}{[\frac{n+1}{n}\cdot \frac{n+2}{n}\cdots \frac{n+n}{n}]}^{1/n}$
$=\underset{n\to \infty }{lim}{[(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n})]}^{1/n}$
$\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}[\ln (1+\frac{1}{n})+\ln (1+\frac{2}{n})+\cdots +\ln (1+\frac{n}{n})]$
$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\ln (1+\frac{r}{n})=\underset{0}{\overset{1}{\int }}\ln (1+x)dx$
Using By-parts, with $\ln (1+x)$ as the first function and $1$ as the second function, we have:
$=\left[x\ln \right(1+x){]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x}{1+x}dx$
$=\ln 2-\underset{0}{\overset{1}{\int }}[1-\left(\frac{1}{1+x}\right)]dx=\ln 2-[x-\ln (1+x){]}_0^1$
$=\ln 2-\left[\right(1-\ln 2)-(0-\ln 1\left)\right]$
$\Rightarrow \ln L=2\ln 2-1=\ln \left(\frac{2^2}{e}\right)$
$\Rightarrow L=\frac{4}{e}$