How to solve this inequation Given two real numbers $0<a<1$ and $0<\delta<1$, I want to find a positive integer $i$ (it is better to a smaller $i$) such that 
$$\frac{a^i}{i!} \le \delta.$$

Here is a not-very-good answer. Let $i$ be the result of rounding $\log\delta/\log a$ up to the nearest integer. Then $i\ge\log\delta/\log a$, so $i\log a\le\log\delta$ (remember, $\log a\lt0$), so $a^i\le\delta$, so $a^i/i!\le\delta$. 
