Finding value of $\lim_{n\rightarrow \infty}\sqrt[n]{\frac{(27)^n(n!)^3}{(3n)!}}$
what i try
$\displaystyle l=\lim_{n\rightarrow \infty}\bigg(\frac{(27)^n(n!)^3}{(3n)!}\bigg)^{\frac{1}{n}}$
$\displaystyle \ln(l)=\lim_{n\rightarrow \infty}\frac{1}{n}\bigg[n\ln(27)+3\ln(n!)-\ln((3n)!)\bigg]$
How do i solve it help me please
By Stirling formula,
$$\sqrt[n]{\frac{(27)^n(n!)^3}{(3n)!}} \sim \left(\frac{(27)^n \left(\sqrt{2\pi n} \left(\frac{n}{e}\right)^n\right)^3}{\sqrt{6\pi n} \left(\frac{3n}{e}\right)^{3n}}\right)^{\frac{1}{n}} = \frac{27 \sqrt{2\pi}^{\frac{3}{n}} n^{\frac{3}{2n}} \left(\frac{n}{e}\right)^3}{\sqrt{6\pi}^{\frac{1}{n}}n^{\frac{1}{2n}}\left( \frac{3}{e}\right)^3 n^3} = \left( \frac{2\pi n}{\sqrt{3}}\right)^{\frac{1}{n}}$$
So the limit you are looking for is equal to $1$.
