Why $\lim_{n \to \infty} \frac{2^nn!}{(2n)!} = 0$ A friend asked me, why
\begin{align}
\lim_{n \to \infty} \frac{2^nn!}{(2n)!} = 0, && n\in \mathbb{N}
\end{align}
and I couldn't answer. We already know that the sequence converges and we are pretty sure that it converges to zero.  And the only thing we are allowed to use is the $\varepsilon$-method.
So let $\varepsilon > 0$. Choose $N:= ?$
\begin{align}
\left|\frac{2^nn!}{(2n)!} - 0 \right| = \frac{2^nn!}{(2n)!} \leq ... < \varepsilon
\end{align}
It seems that I don't know enough about $(2n)!$ so I can't estimate the term. Can you help me?
 A: The expression $$\tag1\frac{(2n)!}{2^nn!}$$
is just the product of the first $n$ odd numbers. To see this note that 
$$2^nn!=(2\cdot2\cdot2\cdot \ldots\cdot 2)\cdot (1\cdot2\cdot 3\cdot \ldots \cdot n) =2\cdot 4\cdot6\cdot\ldots \cdot 2n.$$
The reciprocal of $(1)$ then clearly tends to $0$ as $n\to\infty$.
A: $$\frac{2^n n!}{(2n)!}=\frac{2^n}{\displaystyle\prod_{r=n+1}^{2n} r}<\frac2{2n-1}$$ for $n\ge1$
A: \begin{align}
\lim_{n \to \infty} \frac{2^nn!}{(2n)!} = 0, && n\in \mathbb{N}
\end{align}
\begin{align}
\lim_{n \to \infty} \frac{2^nn!}{(2n).(2n-1).(2n-2).(2n-3).(2n-4).(2n-5)............6.5.4.3.2.1.}= 0, && n\in \mathbb{N}
\end{align}
\begin{align}
\lim_{n \to \infty} \frac{2^nn!}{2^n.n.(n-1).(n-2).(n-3)..............3.2.1.(2n-1).(2n-3).(2n-5)....5.3.1}= 0, && n\in \mathbb{N}
\end{align}
\begin{align}
\lim_{n \to \infty} \frac{2^nn!}{2^n.n!.(2n-1).(2n-3).(2n-5)....5.3.1}= 0, && n\in \mathbb{N}
\end{align}
\begin{align}
\lim_{n \to \infty} \frac{1}{n.(2-\frac{1}{n}).(2-\frac{3}{n}).(2-\frac{5}{n})......................\frac{3}{n}.\frac{1}{n}}= 0, && n\in \mathbb{N}
\end{align}
on applying limit we get the desired result.
A: Alternatively:
$$\frac{2^nn!}{(2n)!} = \frac{2^n}{{2n\choose n}n!}<\frac{2^n}{n!}\to 0$$
A: This follows from the ratio test:
$$\frac{\frac{2^{n+1}(n+1)!}{(2(n+1))!}}{\frac{2^nn!}{(2n)!}}=\frac{2^{n+1}}{2^n}\frac{(n+1)!}{n!}\frac{(2n)!}{(2n+2)!}=\frac 2 1\,\frac{(n+1)}1\,\frac1{(2n+2)(2n+1)}=\frac{1}{2n+1}\ \to\ 0$$
