I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.
Just so you can correct me, my reasoning was:
$$\frac{(3(n+1))!}{(3n)!} = \frac{3\cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdot (…) \cdot 3 \cdot (n+1)}{3 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot (…) \cdot 3 \cdot n } = $$
$$= \frac{3^n(n+1)!}{3^{n}n!} = \frac{(n+1)!}{n!} = n+1$$
Which apparently isn't correct. I must have failed at something very silly. Thanks in advance!
Best Answer
By Stirling's inequality the answer is clearly $\left(\frac{3}{5e}\right)^3$. To prove it, you may notice that by setting $$ a_n = \frac{(3n)!}{(5n)^{3n}} $$ you have: $$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \frac{\frac{3n+3}{5n+5}\cdot\frac{3n+2}{5n+5}\cdot\frac{3n+1}{5n+5}}{\left(1+\frac{1}{n}\right)^{3n}}\to\frac{\left(\frac{3}{5}\right)^3}{e^3}$$ as $n\to +\infty$.