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, $n&gt;0$.

Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$

And I appreciate hint rather than explicit solution.

Thank You.
Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] &lt;\frac{2}{9}\left(m!\right)^3.$$
So you have to go at least far enough so that
$$
\frac{2}{9}\left(m!\right)^3&gt;10^{6},
$$
or $m! &gt; \sqrt[3]{4500000} &gt; 150$.  So $m=5$ (corresponding to $n=15$) isn't far enough; the smallest $n$ satisfying your inequality will be at least $16$.

Similarly, for $n=3m+1$,
$$
3^{-3m-1}(3m+1)!=\left[\left(m+\frac{1}{3}\right)m\left(m-\frac{1}{3}\right)\right]\cdots \left[\frac{4}{3}\cdot1\cdot\frac{2}{3}\right]\cdot\frac{1}{3} &lt; \frac{1}{3}(m!)^3,$$
so you need $m!&gt;\sqrt[3]{3000000}&gt; 140$, and $m=5$ (that is, $n=16$) is still too small.

Finally, for $n=3m+2$,
$$
3^{-3m-2}(3m+2)!=\left[\left(m+\frac{2}{3}\right)\left(m+\frac{1}{3}\right)m\right]\cdots \left[\frac{5}{3}\cdot\frac{4}{3}\cdot1\right]\cdot\frac{2}{3}\cdot\frac{1}{3} &gt; \frac{560}{729}(m!)^3,
$$
where the coefficient comes from the last eight terms, so it is sufficient that $m! &gt; 100\cdot\sqrt[3]{729/560}.$  To show that $m=5$ is large enough, we need to verify that $(12/10)^3=216/125 &gt; 729/560$.  Carrying out the cross-multiplication, you can check without a calculator that $216\cdot 560 =120960$ is larger than $729\cdot 125=91125$, and conclude that $m=5$ (that is, $n=17$) is large enough.  The inequality therefore holds for exactly all $n\ge 17$.
how about make a function?$f(n)=\frac{3^n}{n!}-10^{-6}$ or maybe $f(n)=\frac{3^n}{n!10^{-6}}$