Why $\lim_{n \to \infty} \frac{2^nn!}{(2n)!} = 0$

Question

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?

Answer 1

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 $n\to\infty$.

Answer 2

$$\frac{2^n n!}{(2n)!}=\frac{2^n}{\displaystyle\prod_{r=n+1}^{2n} r}<\frac2{2n-1}$$ for $n\ge1$

Answer 3

\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.