Question

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

$\frac{18}{e^4}$

Let the limit be equal to value A and apply log on both sides.

$\log A=\underset{n\to \infty }{lim}\log {\left(\frac{(n+1)(n+2).....3n}{n^{2n}}\right)}^{\frac{1}{n}}$

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

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{r=2n}\log \left(\frac{n+r}{n}\right)$

Taking $\frac{1}{n}$ as $dx$ and $\frac{r}{n}$ as $x$

Let $a,b$ are the limits of the integration,

$\Rightarrow$ when $r=0$, $a=\underset{n\to \infty }{lim}\frac{1}{n}=0$ ; and $r=2n$, $b=\underset{n\to \infty }{lim}\frac{2n}{n}=2$

$\log A=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{r=2n}\log (1+\frac{r}{n})$

$={\int }_0^2\log (1+x)dx$

$={\left[\right(1+x\left)\log \right(1+x)-(1+x\left)\right]}_0^2$ [$\because \int \log x=x(\log x-1)$]

$\log A=3\log 3-2$

$A=e^{3\log 3-2}$

$=e^{3\log 3}.e^{-2}$

$=\frac{e^{3\log 3}}{e^2}=\frac{e^{{\log }_{e}27}}{e^2}=\frac{27}{e^2}$