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)!?
Best Answer
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?