# Showing that $\frac{x!}{x^{x}}$ tends to zero as x tends to infinity

The question is pretty much in the title, I'm having difficulty formally showing that $\lim\limits _{x\to\infty}\frac{x!}{x^{x}}=0$ (despite intuitively it's fairly obvious).

Thanks in advance for helping ;)

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If $x$ is allowed to take any real value, how do you define $x!$? With the Gamma function? – 1015 Mar 4 '13 at 19:27
Sorry, my bad, I should have specified x takes real values and used the definition with the gamma function. Can I somehow deduce the conclusion for real values directly from it being correct for natural numbers? – Serpahimz Mar 4 '13 at 19:42
To reviewers: The question is not a duplicate of the question about $n!/n^n$. – user53153 Mar 4 '13 at 20:00
Note that the definition is $x!:=\Gamma(x+1)$. – 1015 Mar 4 '13 at 20:24

## 6 Answers

We have that

$$n! \leq \frac{n^n}{2^n} \text{ for } n \geq 6$$

so

$$\lim_{n \to \infty} \frac{n!}{n^n} \leq \lim_{n\to \infty} \frac{1}{2^n} = 0.$$

For $x \in \mathbb{R}$, we know that $\Gamma$ function is increasing for inputs $\geq 2$, so the limit still holds (assume $x > n \geq 6$):

\begin{align} \Gamma(n) &= (n-1)! \\ \Gamma(x) &\leq \Gamma(\lfloor x+1 \rfloor) = \lfloor x \rfloor! \end{align}

so

$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x} \leq \lim_{x \to \infty} \frac{\lfloor x+1 \rfloor!}{\lfloor x \rfloor^{\lfloor x \rfloor}} = \lim_{n \to \infty} \frac{(n+1)!}{n^n} \leq \lim_{n\to \infty} \frac{n+1}{2^n} = 0.$$

I hope this helps ;-)

Edit: Fixed some minor issues, thanks to @julien for noticing.

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If I know the limit holds for naturals I don't think it's correct to directly deduce from continuity it holds for reals, for example $\sin\left(2\pi x\right)$ is constant on the naturals but has no limit in the reals when x tends to infinity. If I knew the limit held for rationals I could use the fact they are dense in the reals to reach the required conclusion. Question is how to show it for rationals? – Serpahimz Mar 4 '13 at 19:49
@Serpahimz It is not correct, fixed now. – dtldarek Mar 4 '13 at 19:52
Oh that's a nice work around. Thanks a lot :) You need another tiny correction though I think since $x!=\Gamma\left(x+1\right)$ and not $\Gamma\left(x\right)$ – Serpahimz Mar 4 '13 at 19:54
Minor detail: $\Gamma$ starts increasing somewhere between $0$ and $1$, but not from $0$, according to the graph in the link you provided. – 1015 Mar 4 '13 at 19:57
Less minor detail: $x!= \Gamma(x+1)$, not $\Gamma(x)$. – 1015 Mar 4 '13 at 20:00

We assume that you want to show that $$\lim_{n\to\infty} \frac{n!}{n^n}=0,$$ where $n$ ranges over the positive integers. For simplicity let $n$ be even, say $n=2m$. Then $$\frac{(2m)!}{(2m)^{2m}}=\frac{m!}{(2m)^m}\cdot \frac{(m+1)(m+2)\cdots(2m)}{(2m)^m}.$$ Note that $\dfrac{m!}{(2m)^m}\le \dfrac{m^m}{(2m)^m}=\dfrac{1}{2^m}$.

Also, $\dfrac{(m+1)(m+2)\cdots(2m)}{(2m)^m}\le 1$.

It follows that $\dfrac{(2m)!}{(2m)^{2m}}\le \dfrac{1}{2^m}$.

A small modification takes care of odd integers. There is a lot of slack.

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So we have $$\frac{x!}{x^x}=\frac{\Gamma(x+1)}{e^{x\ln x}} \qquad \forall x>0.$$

Since $\Gamma(x+1)$ and $x\ln x$ are increasing on $[1,+\infty)$, for $n_x=\lfloor x\rfloor$, we have $$0\leq \frac{\Gamma(x+1)}{e^{x\ln x}}\leq \frac{\Gamma(n_x+2)}{e^{n_x\ln n_x}}=\frac{(n_x+1)!}{n_x^{n_x}}.$$

Now observe that $\lim n_x=+\infty$ as $x$ tends to $+\infty$. So it suffices to show that $$\lim_{n\rightarrow +\infty}\frac{(n+1)!}{n^n}=0.$$

One way (not the best...) to do that is to consider the series $$\sum_{n\geq 1} a_n=\sum_{n\geq 1}\frac{(n+1)!}{n^n}.$$ Since $$\frac{a_{n+1}}{a_n}=\frac{n+2}{n+1}\left( 1-\frac{1}{n}\right)^n\longrightarrow e^{-1}<1,$$ the ratio test tells us the series converges. So the general term tends to $0$, which is what we want.

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Let $\displaystyle u_n=\frac{n!}{n^n}$. By D'Alembert test $$\frac{u_{n+1}}{u_n}=(1+\frac{1}{n})^{-n}=e^{-n\log(1+1/n)}\sim_{+\infty}e^{-1}<1,$$ so the series $\displaystyle \sum_nu_n$ is convergent and consequently: $$\lim\limits _{n\to\infty}\frac{n!}{n^{n}}=0.$$

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It is easy to see that: $$(n!)^2 = (1 \cdot 2 \cdot \ldots \cdot n) \cdot (n \cdot (n - 1) \cdot \ldots \cdot 1)$$ If we substitute all the $n$ factors $n (n -k)$ by their maximum value (which happens at $k^* = \frac{n}{2}$, with value $\frac{n^2}{2^2}$) we have: \begin{align*} (n!)^2 &\le \frac{n^{2n}}{2^{2 n}} \\ n! &\le \left( \frac{n}{2} \right)^n \end{align*} Thus: $$\lim_{n \rightarrow \infty} \frac{n!}{n^n} \le \lim_{n \rightarrow \infty} \frac{n^n}{2^n n^n} = 0$$ Even easier ist just to use Stirling's approximation, but thet crude bound is enough.

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Hint: What is $\log(n!)$? Then replace a sum with integrals

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