Equation with factorial and exponentiation

Question

We know $0 < a \le {1 \over 2}$ and $0 < b \le {1 \over 2}$. I need a good upper
bound for the smallest integer $n \gt 0$ such that:

$$\frac{a^n}{n!} \le b$$

Probably related to the birthday problem,
but motivation is some numerics.

Answer 1

Notice that it is trivial if $a<b$, since

$$\frac{a^1}{1!}<b$$

So the real problem is $0<b<a\le\frac12$. The simplest upper bound:

$$\frac{a^n}{n!}\le a^n\le b\implies n\le\log_a(b)$$

A stronger upper bound using Stirling:

$$\frac{a^n}{n!}\le\frac{(ae)^n}{n^n}\le\left(\frac{ae}2\right)^n\le b$$

$$n\le\log_{ae/2}(b)$$

where $e\approx2.71828$

Answer 2

My favorite simple inequalities for $n!$ are $(n/ei)^n < n! < (n/e)^{n+1}$.

These should help.