Question

$\displaystyle\lim_{n \to\infty} \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. $3 log\,3 - 2$

Answer 1

Answer: (B)

Let $P =\displaystyle\lim_{n \to\infty} \left[\frac{n+1}{n} . \frac{n+2}{n} ... \frac{n+2n}{n}\right]^{\frac{1}{n}}$
Taking log
Log P $ = \displaystyle\lim_{n \to\infty } \frac{1}{n} \displaystyle\sum^{2n}_{r=1} \log\left(1+ \frac{r}{n}\right) $
$= \int\limits^{2}_{0} \log \left(1+x\right)dx $
$= \log\left(1+x\right). x ]^{2}_{0} - \int^{2}_{0} \frac{x}{1+x} dx $
$= 2 \,ln 3 -\int\limits^{2}_{0} \left(1- \frac{1}{1+x}\right)dx = 2In 3 - \left[ x ]^{2}_{0} - ln 3\left(1+x\right) ]^{2}_{0} \right]$
$ = 2 \, ln3 - [2 - 1n3] = 3 ln3 - 2$
$ = ln3^3 - lne^2$
$ = ln \left(\frac{27}{e^2} \right)$
$P = \frac{27}{e^2}$