Question

$\begin{array}{c}lim\\ n\to \infty \end{array}\left(\frac{(n+1)(n+2)...3n}{n^{2n}}\right){}^{\frac{1}{n}}$ is equal to

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

Answer 1

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

Given,

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

$={lim}_{n\to \infty }{\left(\frac{\left(3n\right)!}{n!\cdot n^{2n}}\right)}^{\frac{1}{n}}$

$={lim}_{n\to \infty }{\left(\frac{\sqrt{2\pi \left(3n\right)}{\left(\frac{3n}{e}\right)}^{3n}}{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n\cdot n^{2n}}\right)}^{\frac{1}{n}}$

$={lim}_{n\to \infty }{\sqrt{3}}^{\frac{1}{n}}\cdot \frac{{\left(\frac{3n}{e}\right)}^3}{\left(\frac{n}{e}\right)\cdot n^{2n}}$

$={lim}_{n\to \infty }{\sqrt{3}}^{\frac{1}{n}}\cdot \frac{27}{e^2}$

$=\frac{27}{e^2}$