Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$.
Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$
And I appreciate hint rather than explicit solution.
Thank You.

I would use Stirling's approximation $n!\approx \frac {n^n}{e^n}\sqrt{2 \pi n}$ to get $\left( \frac {3e}n\right)^n \sqrt{2 \pi n} \lt 10^{-6}$.  Then for a first cut, ignore the square root part an set $3e \approx 8$ so we have $\left( \frac 8n \right)^n \lt 10^{-6}$.  Now take the base $10$ log asnd get   $n(\log 8 - \log n) \lt -6$  Knowing that $\log 2 \approx 0.3$, it looks like $16$ will not quite work, as this will become $16(-0.3)\lt 6$.  Each increment of $n$ lowers it by a factor $5$ around here, or a log of $0.7$.  We need a couple of those, so I would look for $18$.
Added:  the square root I ignored is worth a factor of $10$, which is what makes $17$ good enough.a
Alpha shows that $17$ is good enough.
