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

I am not sure if correct but i did it like this :
$(2n)!=(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))\cdot (n!)$ so I have $$\displaystyle \frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}$$ and $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$  is this correct ?  If not why ?
 A: Hint
$$0\leq\frac{n!}{(2n)!}\leq\frac{1}{n}$$
A: 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.
A: Another hint based on using series may be that, if the series $$\sum_0^{\infty}u_n$$ is convergent so $u_n\to 0$. 
A: Hint:
$$ 0 \leq \lim_{n\to \infty}\frac{n!}{(2n)!} \leq \lim_{n\to \infty} \frac{n!}{(n!)^2} = \lim_{k \to \infty, k = n!}\frac{k}{k^2} = \lim_{k \to \infty}\frac{1}{k} = 0.$$
A: If so addressing trivial rigorously I suggest using the notation produtory  to fatorial use the formula $n!=\prod_{k=1}^{n}$ .
\begin{align}
0\leq \frac{n!}{(2n)!}
=
&
\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{2n}k\big)}
\\
=
&
\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=n+1}^{2n}k\big)\big(\prod_{k=1}^{n}k\big)}
\\
=
&
\frac{1}{\big(\prod_{k=n+1}^{2n}k\big)}\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{n}k\big)}
\\
=
&
\frac{1}{\big(\prod_{k=n+1}^{k=2n}k\big)}
\\
=
&
\frac{1}{2n\big(\prod_{k=n+1}^{2n-1}k\big)}
\\
=
&
\frac{1}{2n}\frac{1}{\big(\prod_{k=n+1}^{2n-1}k\big)}
\\
\leq
&
\frac{1}{2n}
\end{align}
