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

Let $y={\left(\frac{\left(n+1\right)\left(n+2\right)\dots . \left(n+2n\right)}{n^{2n}}\right)}^{\frac{1}{n}}$

$y={\left(\frac{\left(n+1\right)}{n}·\frac{\left(n+2\right)}{n}\dots \dots \frac{\left(n+2n\right)}{n}\right)}^{\frac{1}{n}}$

$\log y=\frac{1}{n}\left[\log \left(1+\frac{1}{n}\right)+\log \left(1+\frac{2}{n}\right)+\dots ..\log \left(1+\frac{2n}{n}\right)\right]$

$=\frac{1}{n}\sum _{r=1}^{2n}\ln \left(1+r/n\right)$

Replace $\sum \to \int$ & $\frac{r}{n}\to x$ & $\frac{1}{n}\to dx$

As $n\to \infty$

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

Integrating by parts, we get,

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

$={\left(x·\log \left(1+x\right)\right)}_0^2-\underset{0}{\overset{2}{\int }}\frac{1}{1+x} xdx$

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

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

$=\left(2\log 3-0\right)-\left(2-0\right)+\left(\log 3-\log 1\right)$

$=3\log 3-2$

Since $\log y =3\log 3-2=\log 27-\log e^2$

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

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