# Using Stirling's approximation

I am reading an article and in one section it uses stirling's approximation. I decided to do the math and check if it's ok, but I got a different result than the one in the article.

Where r,p are natural numbers.

I used the "often written" part in Wikipidia (the one with the 'e' in it)

How did the article got this result ?

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We need a direction for a limit I guess. Say, $r$ is fixed but $p \to \infty$. Is that what you want? To tell, we need to know more than just "an article". – GEdgar Mar 25 '12 at 16:47
@GEdgar- you are right, sorry! r is fixed and p tends to infinity – Belgi Mar 25 '12 at 17:13

## 1 Answer

The result mentioned in the paper you are reading holds in the sense that, for every fixed $r\gt2$, $$\lim\limits_{p\to\infty}\frac1p\log{(r-1)p\choose p}=\log c(r),\quad \text{with}\quad c(r)=\frac{(r-1)^{r-1}}{(r-2)^{r-2}}.$$ In other words, when $p\to\infty$, $${(r-1)p\choose p}=c(r)^{p+o(p)}.$$ Stirling's approximation (which you link to) yields the (stronger, non logarithmic) equivalent $${(r-1)p\choose p}=c(r)^p\cdot\frac1{\sqrt{p}}\cdot\sqrt{2\pi}\cdot\sqrt\frac{r-1}{r-2}\cdot(1+o(1)).$$

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What is the base for the log ? if it's p I understand the result in the article. what did you use for your approximation (what did you substitute n! for ?) – Belgi Mar 25 '12 at 17:12
Thank you for the extra explanation! are you sure that it's a small o and not O ? – Belgi Mar 25 '12 at 17:48
Yes.    – Did Mar 25 '12 at 17:53
very interesting. Do you have a reference to the version of Stirling's approximation you are using, or maybe you have another reason for this ? (all versions I know are with big O...) – Belgi Mar 25 '12 at 17:55
Yes: the second formula on the WP page you linked to says that $n!=\sqrt{2\pi n}\cdot(n/\mathrm e)^n\cdot(1+o(1))$. You might want to check again your use of Bachmann-Landau notations. – Did Mar 25 '12 at 18:00