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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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    @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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    oh I see, thanks again!2012-03-25