Is $n! = o(n^n)$?

Question

Intuitively, it seems that $n! = o(n^n)$. We can associate each value in the factorial with a single $n$ to divide by, so $\lim_{n \rightarrow \infty} \frac{n!}{n^n}$ seems to be $0$.

However, I think I have encountered a contradiction. It turns out that $\log(n!) = \Theta(n\log{n})$ (which we can show with Stirling's approximation) or $\log(n!) = \Theta(\log{n^n})$. The log function is strictly increasing, so we can just remove them, so that means that $n! = \Theta(n^n)$, which goes against my earlier hunch of $n! = o(n^n)$.

I must be missing something, right?

$\lim_{n \rightarrow \infty} \frac{n!}{n^n}=0$ does indeed hold. — Michael Greinecker, Feb 01 '14 at 21:08

score 7 · Accepted Answer · answered Feb 01 '14 at 21:14

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$${n!\over n^n}={1\cdot 2\cdots n\over n\cdot n\cdots n}\leq {1\over n}.$$

answered Feb 01 '14 at 21:14

score 4 · Answer 2 · answered Feb 01 '14 at 21:16

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Just because the log function is strictly increasing doesn't mean it preserves asymptotics! For a simple example of this, note that $\log(3^n)=\frac{\log 3}{\log 2}\log(2^n)$, and so $\log(3^n)$ is $\Theta(2^n)$. However, $3^n$ is clearly not $\Theta(2^n)$.

Essentially, if $\log f(n)$ is $\Theta(\log(g(n))$, that means $f(n)/g(n)$ is exponential or smaller.

answered Feb 01 '14 at 21:16

Micah

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Thanks! Why is $\log(3^n)=\frac{\log 3}{\log 2}\log(2^n)$ true? – David Faux Feb 01 '14 at 21:24
By basic properties of logs: $\log(3^n)=n\log(3)$, and $\log(2^n)=n\log(2)$. (Alternately—and equivalently—you could think of it as a version of the change-of-base formula.) – Micah Feb 01 '14 at 21:28

score 3 · Answer 3 · answered Feb 01 '14 at 21:09

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Let $$u_n=\frac{n!}{n^n}$$ then $$\frac{u_{n+1}}{u_n}=\left(1+\frac 1 n\right)^{-n}\to e^{-1}<1$$ so by ratio test the series $\sum u_n$ is convergent so $u_n\to0$ and the desired result follows.

answered Feb 01 '14 at 21:09

score 2 · Answer 4 · answered Feb 01 '14 at 21:09

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Hint: Stirling's approximation states that $$ n! \operatorname*{\sim}_{n\to\infty} \frac{1}{\sqrt{2\pi n}}\frac{n^n}{e^n}$$

so $\frac{n!}{n^n}\xrightarrow[n\to\infty0]{} 0$. Your problem is that you cannot take the exponential of equivalents. For instance, take $f(n)=n$, $g(n)=n+\sqrt(n)$. Clearly, $f(n)\sim g(n)$, yet $$ e^{g(n)} = e^{n+\sqrt{n}} = e^{f(n)}e^{\sqrt n} = \omega\left(e^{f(n)}\right)$$

answered Feb 01 '14 at 21:09

Clement C.

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If you can read French (or make abstraction of the French words around the statements), see this page for what operations you can/cannot do on equivalents. – Clement C. Feb 01 '14 at 21:17

score 1 · Answer 5 · answered Feb 01 '14 at 21:18

If you take the log of Stirling's formula, you get $$\log(n!) \approx n\log n - n + \tfrac{1}{2}\log n + \tfrac{1}{2}\log 2\pi$$ You are keeping only the leading term, $n\log n$, in your question. If you take the difference of the logs (to get handle on the ratio of the original quantities), you get $$\log(n!)-\log(n^n)\approx -n + \text{ lower order terms}$$ Note that as $n\rightarrow\infty$, this does indeed tend to $-\infty$. So the original ratio tends to 0.

Is $n! = o(n^n)$?

5 Answers5