Question

The value of $L=\underset{n\to \infty }{lim}{\left(\frac{(n+1)(n+2)\dots .3n}{n^{2n}}\right)}^{\frac{1}{n}}$ is equal to :

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

Answer 1

The correct option is A $\frac{27}{e^2}$

$\begin{array}{c}\ln L=\frac{1}{n}\left[\ln \left(\frac{n+1}{n}\right)+\ln \left(\frac{n+2}{n}\right)+........\ln \left(\frac{n+2n}{n}\right)\right]\\ \ln L=\frac{1}{n}\left[\ln \left(1+\frac{1}{n}\right)+\ln \left(1+\frac{2}{n}\right)+........\ln \left(1+\frac{2n}{n}\right)\right]\\ \ln L=\frac{1}{n}{\sum }_{r=1}^{2n}\ln \left(1+\frac{r}{n}\right)\\ \ln L={\int }_0^2\ln \left(1+x\right)dx\\ 1+x=t\\ \ln L={\int }_1^3\ln tdt\\ \ln L={\left[t\ln t-t\right]}_1^3\\ =\left(3\ln 3-3\right)-\left(0-1\right)\\ \ln L=3\ln 3-2=\ln \left(\frac{27}{e^2}\right)\\ L=\frac{27}{e^2}\\ Hence,\\ optionAiscorrectanswer.\end{array}$