Find the limit $\lim_{n\to\infty} \frac{(3n)! \, e^n}{(2n)!\,n^n\,8^n}$

Question

Find $$\lim_{n\to\infty} \frac{(3n)! \, e^n}{(2n)!\,n^n\,8^n}$$

I tried by simplifying $n!$ divided by $n! = 1$? What should I do next? I get then $3!e^n / 2!n^n8^n$

Answer 1

HINT $$ \frac{(3n)! e^n}{(2n)! n^n 8^n} = \left(\frac{e}{8n}\right)^n \prod_{k=1}^n (2n+k) = \prod_{k=1}^n \frac{2n+k}{8n/e} = \prod_{k=1}^n \left(\frac{e}{4} + \frac{ke}{8n}\right) $$

Answer 2

Alternative Method:

Obviously, $$0\le\frac{(3n)! e^n}{(2n)! n^n 8^n}$$

Also, observe $$\frac{(3n)! e^n}{(2n)! n^n 8^n}=\frac{(3n)!}{(2n)!}\cdot(\frac{e}{8n})^n\le\frac{(3n)^n}{n^n}\cdot(\frac{e}{8n})^n=(\frac{3e}{8n})^n$$

So $$0\le\frac{(3n)! e^n}{(2n)! n^n 8^n}\le(\frac{3e}{8n})^n$$ $$\lim_{n\to\infty}0\le\lim_{n\to\infty}\frac{(3n)! e^n}{(2n)! n^n 8^n}\le\lim_{n\to\infty}(\frac{3e}{8n})^n$$ $$0\le\lim_{n\to\infty}\frac{(3n)! e^n}{(2n)! n^n 8^n}\le0$$

So by Squeeze Theorem,

$$\lim_{n\to\infty}\frac{(3n)! e^n}{(2n)! n^n 8^n}=0$$