_ n → ∞ ( (n + 1)(n + 2) . 3n n ^ 2n )(1)/(n) is equal to

$\lim_{n \to \infty}\left(\dfrac{\left(n + 1\right)\left(n + 2\right) \ldots . 3n}{{n}^{2n}}\right)\dfrac{1}{n}$

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

$y = \left(\dfrac{\left(n + 1\right)}{n} \cdot \dfrac{\left(n + 2\right)}{n} \ldots \ldots \dfrac{\left(n + 2n\right)}{n}\right)\dfrac{1}{n}$

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

$= \dfrac{1}{n}\mathop{∑}_{r = 1}^{2n}\ln \left(1 + r / n\right)$

Replace $∑ \to ∫$ & $\dfrac{r}{n} \to x$ & $\dfrac{1}{n} \to dx$

As $n \to \infty$

$\log y = \mathop{∫}_{0}^{2}\log \left(1 + x\right)dx$

Integrating by parts, we get,

$\log y = \mathop{∫}_{0}^{2}1 \cdot \log \left(1 + x\right)dx$

$= {\left(x \cdot \log \left(1 + x\right)\right)}_{0}^{2} - \mathop{∫}_{0}^{2}\dfrac{1}{1 + x}xdx$

$= {\left(xlog\left(1 + x\right)\right)}_{0}^{2} - \mathop{∫}_{0}^{2}\left(\dfrac{1 + x}{1 + x}\right)dx + \mathop{∫}_{0}^{2}\dfrac{1}{1 + x}dx$

$= {\left(xlog\left(1 + x\right)\right)}_{0}^{2} - {\left(x\right)}_{0}^{2} + {\left(\log \left(1 + x\right)\right)}_{0}^{2}$

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

$= 3\log 3 - 2$

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

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

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