Find the smallest $n$ such that $\frac{a^n}{n!} < \epsilon $ For $a>0$ and $n\in\mathbb{N}$. I think the archimedean property of the reals tells us that for $n$ fixed, there exists some $\epsilon > 0$ such that the inequality is true. But how can you choose $n$ for fixed $\epsilon$?
Context: Trying to show that the sequence $\{ c_n = a^n/n! \}$ converges to zero, that is to say, for any fixed $\epsilon$, there's some $N$ such that $c_n < \epsilon$ for all $n \geq N$. How do find that $N$?
Edit: Another idea: Show that the sequence is Cauchy (so it must converge in $\mathbb{R}$); then suppose it converges to something other than 0; expose a contradiction.
Edit: Found another way to prove that the sequence (see Context) converges, so the question is void. (Just write $\{c_n\}$ as the product of $\{1,c_1,c_2,...\}$ and $\{a,a/2,a/3,...\}$; the former is bounded and the latter converges to 0, so the product rule for limits tells us that $c_n \rightarrow 0$.)
 A: From a purely algebraic point of view, consider the equation $$a^n=\epsilon \times n!$$ Take the logarithms of both sides and the equation becomes $$n\log(a)=\log(\epsilon)+\log(n!)$$ Now use Stirling approximation $$\log(n!)\approx \frac{1}{2}\log(2\pi)+(n+\frac{1}{2})\log(n)-n$$ so you have the equation to solve. Use any graphing tool and you are done.
For example, using $a=\frac{3}{4}$,$\epsilon=10^{-20}$ we find $n=19.3826$ while the exact solution is $n=19.3812$
A: You won't get much further by using elementary ideas. The trick is to prove that the factorial grows faster than an exponential, and this is somehow hard. If you don't want to use Stirling, you can proceed as follows. Define
$$
p_n = \frac{a^n}{n!}
$$
and notice that
$$
\frac{p_{n+1}}{p_n} = \frac{a}{n}.
$$
Then
$$
p_{2} = a p_1, \quad p_3 = \frac{a}{2}p_2 = \frac{a}{2} a p_1 = \frac{a^2}{2}p_1,
$$
and similarly
$$
p_4 = \frac{a}{4}p_3 = \frac{a^3}{2^3}p_1, \ldots
$$
You should now guess that $p_n \approx 2^{-n+1}p_1$, and therefore $p_n \to 0$. This will not provide you with the smallest $n$ for a given $\varepsilon$, although you can get an estimate. But please remark that you don't need the smallest $n$, to prove that the limit is zero.
