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

$\frac{9}{e^2}$

Let $A=\lim\limits_{n \rightarrow \infty}\left\{\frac{(n+1)(n+2)(n+3) ... . .3 n}{n^{2 n}}\right\}^{\frac{1}{n}}$

Then,

$\log A =\lim\limits_{n \rightarrow \infty} \frac{1}{n} \log \left\{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right) ...\left(1+\frac{2 n}{n}\right)\right\}$

$\Rightarrow \log A =\lim\limits_{n \rightarrow \infty} \frac{1}{n} \sum\limits_{r=1}^{2 n} \log \left(1+\frac{r}{n}\right)$

$\Rightarrow \log A=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^{2 n} \log \left(1+\frac{r}{n}\right) . \frac{1}{n}$

$\Rightarrow \log A=\int\limits_0^2 \log (1+x) d x$

$\Rightarrow \log A=\int\limits_0^2 \log (1+x) . 1 d x$

$\Rightarrow \log A=[x \log (1+x)]_0^2-\int\limits_0^2 \frac{x}{1+x} d x$

$\Rightarrow \log A=2 \log 3-[x-\log (1+x)]_0^2$

$\Rightarrow \log A=2 \log 3-(2-\log 3)=3 \log 3-2$

$\Rightarrow \log A=\log 27-\log e^2=\log \left(\frac{27}{e^2}\right)$

$\Rightarrow A=\frac{9}{e^2}$