Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

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

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

$=\underset{0}{\overset{2}{\int }}log(1+x)dx$

$=\left[log\right(1+x)dx{]}_0^2-\underset{0}{\overset{2}{\int }}\frac{1.x}{1+x}dx$

= 2 log 3 - $\left[\underset{0}{\overset{2}{\int }}(1-\frac{1}{1+x})dx\right]$

= 2 log 3 - $[x-log(1+x){]}_0^2$

= 2 log3 – (2 – log3)

log p = 3log 3 – 2

p = $e^{3log3-2}=\frac{e^{log27}}{e^2}=\frac{27}{e^2}$