Rate of convergence of $\frac{a^n}{n!}$ It is a well-known result from calculus that
$$ \lim_{n \to \infty} \frac{a^n}{n!} =0,$$
for every $a \in \mathbb{R}$. However, the classical proof involves proving the convergence of the infinite series
$$ \sum_{n=1}^{\infty} \frac{a^n}{n!}.$$
I am working on some numerical algorithms at the moment. I am wondering if, for every $\epsilon>0$, we can construct an $\mathbf{\text{explicit formula}}$ for $n \in \mathbb{N}$ (in terms of $a$ and $\epsilon$) such that
$$ \bigg| \frac{a^n}{n!} \bigg| < \epsilon.$$
Do I need to use any special inequality for this? Any ideas? 
 A: By Stirling's Approximation for $n!$,
$$\frac{|a|^n}{n!}<\frac{|a|^n}{\sqrt{2\pi}n^{n+1/2}e^{-n}}<\frac{|a|^n}{\sqrt{2\pi}n^{n}e^{-n}}$$ so it is sufficient to solve
$$\begin{align}
\frac{|a|^n}{\sqrt{2\pi}n^{n}e^{-n}}
&<\varepsilon\\
\frac{|a|^ne^{n}}{n^{n}}
&<\sqrt{2\pi}\varepsilon\\
\left(\frac{|a|e}{n}\right)^n
&<\sqrt{2\pi}\varepsilon\\
\frac{1}{\sqrt{2\pi}\varepsilon}
&<\left(\frac{n}{|a|e}\right)^n\\
\left(\frac{1}{\sqrt{2\pi}\varepsilon}\right)^{\frac{1}{|a|e}}
&<\left(\frac{n}{|a|e}\right)^{\frac{n}{|a|e}}\\
-\frac{1}{|a|e}\ln\left(\sqrt{2\pi}\varepsilon\right)
&<\frac{n}{|a|e}\ln\left(\frac{n}{|a|e}\right)\\
W\left(-\frac{1}{|a|e}\ln\left(\sqrt{2\pi}\varepsilon\right)\right)
&<W\left(\frac{n}{|a|e}\ln\left(\frac{n}{|a|e}\right)\right)\\
W\left(-\frac{1}{|a|e}\ln\left(\sqrt{2\pi}\varepsilon\right)\right)
&<\ln\left(\frac{n}{|a|e}\right)
\end{align}$$
where $W$ is the Lambert function. So it suffices to take $$n=\left\lceil |a|e\cdot e^{W\left(-\frac{1}{|a|e}\ln\left(\sqrt{2\pi}\varepsilon\right)\right)}\right\rceil$$
For example, with $a=100$, $\varepsilon=0.01$, this gives $n=276$. Direct computation (with a computer) of $\frac{100^{276}}{276!}\approx0.00035$ shows this is only a tad smaller than $0.01$. Trial and error with a computer's help reveals that the critical $n$ is $273$. This is evidence that the bound is pretty close to sharp.
(Computing values of $W$ is beyond most simple computing devices. But a Computer Algebra System can handle it. At WolframAlpha for example, you can use LambertW(...).)
A: Fix $a$.  Then, take $n>2a^2$.  Therefore, we have
$$\begin{align}
\frac{a^n}{n!}&\le \frac{a^n}{(n/2)^{n/2}}\\\\
&=\left(\frac{a}{\sqrt{n/2}}\right)^n\\\\
&\le \frac{a}{\sqrt{n/2}}\\\\
&<\epsilon
\end{align}$$
whenever $n>2a^2\max\left(1,\frac{1}{\epsilon^2}\right)$.
