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Let x be a complex number.

What is the Stirling formula for x(x+1)(x+2)...(x+n-1) when n goes to infinity?

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It looks like you want a formula for the asymptotics of the Pochhammer symbol $(x)_n$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:

$$(x)_n \sim \frac{2\pi}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$

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    $\begingroup$ I would say that equivalent was obtained through the generalized Stirling formula for the $\Gamma$ function, as pointed to by Yemon Choi. $\endgroup$ Mar 12, 2011 at 15:20
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    $\begingroup$ That seems to be a reasonably likely explanation. I just Googled for Pochhammer asymptotic, and didn't bother to think too much. $\endgroup$
    – S. Carnahan
    Mar 12, 2011 at 15:43

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