Which function grows faster: e^n or (log n) factorial?

Question

How would I solve:

e^n / (log n)!

where the limit n→infinity

I know that n! grows faster than e^n.

I think that (log n)! would grow faster than e^n because the value of log n would eventually be close to the value of n that I mentioned above, if that makes sense. Is my reasoning correct? What would be the bounds for a function like (log n)!?

Answer 1

You could let $m = \log n,$ that is, $n = e^m,$ so that $m \to \infty$ as $n \to \infty,$ and therefore

$$ \lim_{n\to\infty} \frac{e^n}{(\log n)!} = \lim_{m\to\infty} \frac{e^{e^m}}{m!}. $$

Let $f(m) = e^{e^m}/m!.$ Then $$ \frac{f(m+1)}{f(m)} = \frac{e^{e^{m+1}}/(m+1)!}{e^{e^m}/m!} = \frac{e^{(e-1)e^m}}{m+1}. $$

Can you figure it out from here?

Answer 2

let's calculate the derivative for them.

(e^n)' = e^n = a
(log(n))' = 1/(n*ln(2)) = b

If n increases, then a increases and b decreases. So, e^n grows faster.

Answer 3

Taking logs of both terms: $\mathrm{log}(e^n) = n$ and $\mathrm{log}((\mathrm{log}\ n)!) = (\mathrm{log}\ n)(\mathrm{log}\mathrm{log}\ n) - \mathrm{log}\ n + O(\mathrm{log}\mathrm{log}\ n)$. (Using Stirling's approximation $\mathrm{log}\ k! = k \mathrm{log}\ k - k + O(\mathrm{log}\ k)$).

For $n>3$, $\mathrm{log}\mathrm{log}\ n < \mathrm{log}\ n$, so $\mathrm{log}((\mathrm{log}\ n)!) < 2 (\mathrm{log}\ n)^2$ (for large enough $n$).

All powers of $\mathrm{log}\ n$ are o($n$), so $\log(e^n)$ grows faster than $\mathrm{log}((\mathrm{log}\ n)!)$. And so $e^n$ also grows faster than $(\mathrm{log}\ n)!$