A curious limit

**SworD** · September 2nd 2012, 10:28 PM

I stumbled upon this somewhere but didn't know how to prove it.

$\lim_{x\to\infty}\frac{e^x \cdot x!}{\sqrt{x} \cdot x^x}= \sqrt{2\pi}$

Anyone have any ideas why this is the case or how to derive it? I suspect it has something to do with the gamma function.

**JJacquelin** · September 2nd 2012, 10:47 PM

The Stirling's approximation of x! for large x is :
(x^x)*exp(-x)*sqrt((2x+(1/3))*pi) equivalent to (x^x)*exp(-x)*sqrt(2*pi)*sqrt(x)

**SworD** · September 2nd 2012, 10:51 PM

Ah, yeah when I look at that closely it works. Thanks

**MaxJasper** · September 2nd 2012, 10:55 PM

Originally Posted by SworD

I stumbled upon this somewhere but didn't know how to prove it.

$\lim_{x\to\infty}\frac{e^x \cdot x!}{\sqrt{x} \cdot x^x}= \sqrt{2\pi}$

Anyone have any ideas why this is the case or how to derive it? I suspect it has something to do with the gamma function.

Expand the argument to its Taylor series and then let $x\to \infty$

$\lim_{x\to \infty } \, \frac{e^x x!}{\sqrt{x} x^x}$ = $\lim_{x\to\infty} \sqrt{2 \pi }+\frac{\sqrt{\frac{\pi }{2}}}{6 x}+\frac{\sqrt{\frac{\pi }{2}}}{144 x^2}-\frac{139 \sqrt{\frac{\pi }{2}}}{25920 x^3}-\frac{571 \sqrt{\frac{\pi }{2}}}{1244160 x^4}+\frac{163879 \sqrt{\frac{\pi }{2}}}{104509440 x^5}+O\left(\left(\frac{1}{x}\right)^6\right)$ = $\sqrt{2 \pi }$

**SworD** · September 2nd 2012, 11:04 PM

o.O Could you elaborate on how you arrived at that series? Unless you are extending JJacquelin's answer by inputting Stirling's formula then I'm confused.. and isnt this rather a Laurent series?

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