Question

$\underset{n\to \infty }{\lim }{\left(\frac{\left(n+1\right)\left(n+2\right)....3n}{n^{2n}}\right)}^{\frac{1}{n}}$ is equal to :

• A. $\frac{18}{e^4}$
• B. $\frac{27}{e^2}$
• C. $\frac{9}{e^2}$
• D. $3log3-2$

Answer 1

Answer: (B)

Let $P=\underset{n\to \infty }{\lim }{\left[\frac{n+1}{n}.\frac{n+2}{n}...\frac{n+2n}{n}\right]}^{\frac{1}{n}}$
Taking log
Log P $=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^{2n}\log \left(1+\frac{r}{n}\right)$
$=\underset{0}{\overset{2}{\int }}\log \left(1+x\right)dx$
$=\log \left(1+x\right).x{]}_0^2-{\int }_0^2\frac{x}{1+x}dx$
$=2ln3-\underset{0}{\overset{2}{\int }}\left(1-\frac{1}{1+x}\right)dx=2In3-\left[x{]}_0^2-ln3\left(1+x\right){]}_0^2\right]$
$=2ln3-[2-1n3]=3ln3-2$
$=ln{3}^3-ln{e}^2$
$=ln\left(\frac{27}{e^2}\right)$
$P=\frac{27}{e^2}$