Show that $\lim \limits_{n\rightarrow\infty}\frac{n!}{(2n)!}=0$

Hint

$$0\leq\frac{n!}{(2n)!}\leq\frac{1}{n}$$

It's correct, but I imagine you're expected to show a bit more work to justify your assertion that $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$ An easy way to do this is to bound this sequence of fractions with another, simpler one whose limit you know is 0.

Another hint based on using series may be that, if the series $$\sum_0^{\infty}u_n$$ is convergent so $u_n\to 0$.