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An article I'm reading states a bound on some value, I believe they are using Stirling's approximation but I have been unable to derive their result, perhaps someone here can help.

The claim:

For $\alpha,\epsilon >0$ we have $$ \frac{\alpha^n}{n!}<\epsilon $$ when $$ n>e\alpha + \log(1/\epsilon). $$

My attempt so far:

By Stirling's approximation we have $$ \frac{\alpha^n}{n!}\sim\frac{e^n\alpha^n}{\sqrt{2\pi n}\;n^n}<\frac{e^n\alpha^n}{n^n}, $$ inserting into the first inequality and taking the log of both sides we have $$ n\log(e\alpha/n) < \log\epsilon $$ then, taking the inverse of the log arguments and rearranging we have $$ n\log n > n \log e\alpha + \log(1/\epsilon). $$ I am unsure how to proceed, is there another approximation I should be using?

Duong Ngo
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DeafIdiotGod
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2 Answers2

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This community wiki is a rewrite of my comment. By doing so, the question will be no longer unanswered.

$n > e\alpha + \log(1/\epsilon) \Longleftrightarrow \log(\epsilon) > e\alpha - n \Longleftrightarrow \epsilon > \dfrac{e^{e\alpha}}{e^{n}}$. To prove that $\dfrac{\alpha^{n}}{n!} < \epsilon$, we will show that $\dfrac{\alpha^{n}}{n!} < \dfrac{e^{e\alpha}}{e^{n}}$, which is equivalent to $\dfrac{{(e\alpha)}^{n}}{n!} < e^{e\alpha}$. However, this follows immediately from the inequality $\dfrac{x^{n}}{n!} < e^{x}$ for $x > 0$ (because $e^{x} = \displaystyle\sum^{\infty}_{k=0}\dfrac{x^{k}}{k!}$) when we substitute $x = e\alpha$.

Duong Ngo
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$$\frac{\alpha^n}{n!}<\epsilon\quad \implies \quad n!>\frac {\alpha^n}\epsilon$$ Consider the equality and have a look here. @robjohn gave the superb approximation $$n \sim e\alpha \,\exp\Bigg(W\left(-\frac{1}{2 e \alpha }\log \left(2 \pi \alpha \epsilon ^2\right)\right) \Bigg)-\frac 12$$ where $W(.)$ is the principal branch of Lambert function.

Just to give an idea, suppose $\alpha=\pi$ and $\epsilon=10^{-6}$; this will give as a real $n\sim 16.8649$ while the "exact" solution is $16.8663$.

Now, you could use as a very first approximation $$W(x) \sim \log (x)-\log (\log (x))\quad \implies \quad e^{W(x)}\sim \frac{x}{\log (x)}$$

Just continue.

Claude Leibovici
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