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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.

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    Are $a,b$ positive? – 2017-02-16

2 Answers 2

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Notice that it is trivial if $a

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

So the real problem is $0

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

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 $e\approx2.71828$

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    Works also for a >= 1/2 I guess! Like for example a < 2/e. – 2017-02-16
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    Of course, Stirling's approximation works fine for larger $n$. – 2017-02-16
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My favorite simple inequalities for $n!$ are $(n/ei)^n < n! < (n/e)^{n+1}$.

These should help.

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    It's pretty hard to solve for $n$ in things like $(cn)^n=b$, so you may wish to elaborate. – 2017-02-16
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    Possibly small typo ei -> e. – 2017-02-16
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    @j4nbur53 I thought the same, but that's kind of the right side of the inequality? Maybe? not entirely sure what's going on here. – 2017-02-16