If $f(n) =\frac{1}{n} [(n + 1) (n + 2)(n + 3) . (n + n)]^{1/n}$ then $\underset{n→∞}{\lim}f(n)$ equals

$4/e$

$A=\underset{n→∞}{\lim}\frac{1}{n}[(n + 1)(n + 2)(n + 3)...(n + n)]^{1/n}$

$=\underset{n→∞}{\lim}\left[(1+\frac{1}{n})+(1+\frac{2}{n})+(1+\frac{3}{n})+...(1+\frac{n}{n})\right]^{1/n}$

$\log A =\underset{n→∞}{\lim}\frac{1}{n}\sum\limits_{r=1}^{n}\log(1+\frac{r}{n})$

$=\int\limits_{0}^{1}\log(1+x)dx⇒A=\frac{4}{e}$