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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?

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  • $\begingroup$ Take logarithms and use Stirling approximation for log(n!) $\endgroup$ Feb 2 '14 at 15:13
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    $\begingroup$ Suggestion: For full absolute value signs, use left| and right| instead of \Big. I usually save \Big for sets, topology, and such. $\endgroup$ Feb 2 '14 at 15:19
  • $\begingroup$ Intuitively this is obvious. If you take $n$ pair of twins and line those $2n$ people up randomly, the probability that everyone ends up next to his/her twin (necessarily in pairs $(2i-1,2i)$ of positions) clearly goes to zero as $n\to\infty$. $\endgroup$ Feb 2 '14 at 15:23
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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 $n\to\infty$.

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

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  • $\begingroup$ It beats me why you should estimate the product of factors${}<1$ by the second-smallest factor rather than by the smallest factors: $\frac2{2n}=\frac1n$ $\endgroup$ Feb 2 '14 at 16:07
  • $\begingroup$ @MarcvanLeeuwen, Initially, I was working as Hagen did in his answer later. Its still valid as $<\frac2{2n},$ right ? $\endgroup$ Feb 2 '14 at 16:18
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\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.

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Alternatively: $$\frac{2^nn!}{(2n)!} = \frac{2^n}{{2n\choose n}n!}<\frac{2^n}{n!}\to 0$$

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This follows from the ratio test: $$\frac{\frac{2^{n+1}(n+1)!}{(2(n+1))!}}{\frac{2^nn!}{(2n)!}}=\frac{2^{n+1}}{2^n}\frac{(n+1)!}{n!}\frac{(2n)!}{(2n+2)!}=\frac 2 1\,\frac{(n+1)}1\,\frac1{(2n+2)(2n+1)}=\frac{1}{2n+1}\ \to\ 0$$

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