Question

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

Answer: (A)

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

$L=\underset{n\to \infty }{lim}{((1+\frac{1}{n})\dots ..(1+\frac{2n}{n}))}^{1/n}$

$\log L=\underset{n\to \infty }{lim}\frac{1}{n}\log (1+\frac{1}{n})+\dots .+\log (1+\frac{2n}{n})$

$=\underset{n\to \infty }{lim}\frac{1}{n}\log (1+\frac{1}{n})+\frac{1}{n}\log (1+\frac{2}{n})+...\dots +\frac{1}{n}\log (1+\frac{2n}{n})$

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

$=\log (1+x).x{|}_o^2-{\int }_o^2\frac{x}{x+1}dx$

$=\left[\log \right(1+x).x-x+\log (x+1){]}_o^2$

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

$\Rightarrow L=\frac{27}{e^2}$.