Question

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

Answer 1

Solution

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

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

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

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

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

$logL=log(1+x)x-\int \frac{x}{x+1}dx$

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

$logL=2log3-2+log3=log\left(3^2\right)-log\left(e^2\right)+lo{g}^3$

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