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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.

A screenshot of the use of stirling's approximation

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".2012-03-25
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    @GEdgar- you are right, sorry! r is fixed and p tends to infinity2012-03-25

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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 ?)2012-03-25
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    Thank you for the extra explanation! are you sure that it's a small o and not O ?2012-03-25
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    Yes. $ $ $ $ $ $2012-03-25
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    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...)2012-03-25
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    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](http://en.wikipedia.org/wiki/Big_O_notation).2012-03-25
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    oh I see, thanks again!2012-03-25