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  1. #1
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    A curious limit

    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.
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  2. #2
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    Re: A curious limit

    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)
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    Re: A curious limit

    Ah, yeah when I look at that closely it works. Thanks
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    Lightbulb Re: A curious limit

    Quote Originally Posted by SworD View Post
    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 }
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    Re: A curious limit

    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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