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By definition, for any $a\in\mathbb R$, there exists $k\in\mathbb N^*$ such that, if $n\in\mathbb N+k$, then $n!^{\large\frac1n}>a$. Therefore, by induction, I must:

  • find some $k$ that satisfies $k!>a^k$;
  • show that, for each $n\ge k$, if $n!>a^n$, then $(n+1)!>a^{n+1}$.

For the 2nd part, it's enough that $k\ge a-1$. I noticed the smallest adequate $k$ is $4$ for $a=2$ and $7$ for $a=3$, and I'm convinced that one can always find a $k$, but I would like a function of $a$ to make an actual argument.

Definition aside, methods other than this one are welcome. I tried picking a sequence dominated by $\big(n!^{\large\frac1n}\big)$ that went to infinity as well, but couldn't find any.

Luke
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6 Answers6

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When $n$ is even, $n/2$ of the numbers in the sequence $1,\ldots,n$ are greater than $n/2$. Therefore, $n! \geq (n/2)^{n/2}$, and this leads to the desired result.

Lord Soth
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  • Simple and clear. I was about to post this, until I saw it had already been. This should be the accepted answer, in my opinion. Note that it's also true that $n^{n/2} < n! ⇒ \sqrt{n} < n!^{1/n}$. – GregRos Mar 29 '14 at 13:35
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Start with the logarithm of what we want $$ \begin{align} \frac1n\log(n!) &=\frac1n\sum_{k=1}^n\log(k)\\ &\ge\frac1n\int_1^n\log(x)\,\mathrm{d}x\\ &=\log(n)-1+\frac1n \end{align} $$ Therefore, $$ \left(n!\right)^{1/n}\ge\frac ne $$

robjohn
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Using AM-GM:

$$\frac{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}}{n} \ge \sqrt[n]{\frac{1}{n!}}$$ $$\sqrt[n]{n!} \ge \frac{n}{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}} = \frac{n}{H_n}$$

But since $H_n$ grows as $\ln x$ , we have (as $n\to\infty$):

$$\frac{n}{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}} \to \infty$$

and therefore (as $n\to\infty$),

$$\sqrt[n]{n!} \to \infty$$

Yiyuan Lee
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Using Stirling's approximation, we have $n!> \sqrt{2\pi n}(n/e)^n$. Then $(n!)^{1/n}> \sqrt[n]{2\pi}n^{1/n} \frac{n}{e}$. Since $\sqrt[n]{2\pi}\rightarrow 1$ and $n^{1/n}\rightarrow 1$ but $\frac{n}{e}\rightarrow \infty$, we see that $(n!)^{1/n}\rightarrow \infty$.

Hayden
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Here is a simple but not-so-elegant proof.

Without loss, it is enough to prove the result for fixed natural a > 1. Showing that $(n!)^{1/n} > a$ eventually(ev.) is the same as showing $$n! > a^n \quad\text{ev.} \iff x_n := \frac{n!}{a^n} > 1\quad\text{ev.}$$

For $n>a$,

$$x_n = \frac{a!}{a^a} \frac{n(n-1)…(a+1)}{a^{n-a}} $$

Every term on the top of the second fraction is at least a+1, and there are $n-(a+1)+1 = n-a$ terms, so

$$ x_n \geq \frac{a!}{a^a} \frac{(a+1)^{n-a}}{a^{n-a}} = \frac{a!}{a^a} \left( \frac{a+1}{a} \right)^{n-a} = C(a) \left( \frac{a+1}{a} \right)^{n-a} $$ ($C(a) = \frac{a!}{a^a} > 0$ is constant in $n$) which is bigger than 1 eventually, hence the result.

Calvin Khor
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  • Makes sense, but that $a^{n−a}$ shouldn't be there, should it? I edited it out, feel free to rollback if needed.

    Also, you assume $a\in\mathbb N$, but that's without loss too; one just has to replace $a$ with $\lfloor a\rfloor$.

     – Luke Mar 29 '14 at 05:48
  • I think what you want is $$x_n=\frac{a!}{a^{\color{#C00000}{a}}}\frac{n(n-1)\dots(a+1)}{a^{n-a}}$$ then $$x_n\ge C(a)\left(\frac{a+1}{a}\right)^{n\color{#C00000}{-a}}$$ – robjohn Mar 29 '14 at 05:55
  • thanks for the corrections, I've edited them in. I wanted to hide the factor $\left(\frac{a+1}{a}\right)^{-a}$ in $\left(\frac{a+1}{a}\right)^{n-a}$ into $C(a)$ but I suppose that is not as clear as I thought. – Calvin Khor Mar 29 '14 at 12:35
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    You could do that with $$\frac{a!}{a^a}\left(\frac{a+1}{a}\right)^{n-a}=\frac{a!}{(a+1)^a}\left(\frac{a+1}{a}\right)^n$$ – robjohn Mar 29 '14 at 13:00
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You have to prove: $$ \forall a > 0 \, \exists n \, \ni \, n! > a^n $$ which is equivalent to $$ \forall a > 0 \, \exists n \, \ni \, \frac{a^n}{n!} < 1 \, . $$ This last statement follows from the fact that $$ e^a = \sum_{n = 0}^\infty \frac {a^n}{n!} $$ is convergent for all $ a > 0$.

Hans Engler
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