Why is a double exponential function faster than $x!$?
By Daniel Johnston
Reading this wikipedia article I found an interesting sentence:
Factorials grow faster than exponential functions, but much slower than double-exponential functions.
The author doesn't provide a link let alone a proof of that fact and I don't find it obvious. After all, factorial functions grow really fast but the author says they grow much slower than double exponentials. Does anyone know a proof that
$$\lim_{x \to \infty} \frac{e^{e^{x}}}{x!} = \infty$$
$\endgroup$ 23 Answers
$\begingroup$Simple comparison of factors shows that $$ n!=\prod_{k=1}^nk\le\prod_{k=1}^nn =n^n\tag{1} $$ Since $$ \begin{align} \lim_{n\to\infty}\left(e^n-n\log(n)\right) &=\lim_{n\to\infty}e^n\left(1-\frac{n\log(n)}{e^n}\right)\\[6pt] &=\infty\cdot1\tag{2} \end{align} $$ we have $$ \begin{align} \lim_{n\to\infty}\frac{e^{e^n}}{n^n} &=\lim_{n\to\infty}e^{e^n-n\log(n)}\\[6pt] &=\infty\tag{3} \end{align} $$ Applying $(1)$ to $(3)$ gives $$ \lim_{n\to\infty}\frac{e^{e^n}}{n!}=\infty\tag{4} $$
$\endgroup$ 0 $\begingroup$Don't need Stirling for that. It is much more basic:
Let $a_n := \frac{e^{e^n}}{n!}$. Then $$\frac{a_{n+1}}{a_{n}} = \frac{e^{e^{n}e}}{e^{e^n}(n+1)} =\frac{1}{n+1}\frac{(e^{e^n})^e}{e^{e^n}} = \frac{1}{n+1}(e^{e^n})^{e-1} \geq \frac{e^{n(e-1)}}{n+1}$$ Now you only need to know that $\exp$ grows alot faster than each polynomial. This gives you that even $\frac{a_{n+1}}{a_n}$ grows really fast.
$\endgroup$ $\begingroup$The Stirling formula gives $\ln(x!)$ is about $x \ln(x) - x$ for large $x$ so $$ \ln\left(\frac{e^{e^x}}{x!}\right) = e^x - \ln(x!) \approx e^x - x \ln(x) + x \to \infty $$
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