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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3$\begingroup$ I may be missing something, but how about en.wikipedia.org/wiki/… $\endgroup$– Yemon ChoiCommented Mar 12, 2011 at 5:08
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1 Answer
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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:
$$(x)_n \sim \frac{\sqrt{2\pi}}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$
answered Mar 12, 2011 at 5:48
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1$\begingroup$ I would say that equivalent was obtained through the generalized Stirling formula for the $\Gamma$ function, as pointed to by Yemon Choi. $\endgroup$ Commented Mar 12, 2011 at 15:20
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1$\begingroup$ 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 ♦Commented Mar 12, 2011 at 15:43
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$\begingroup$ Shouldn't this be $\sqrt{2\pi}$ in the numerator? $\endgroup$ Commented Jun 29 at 8:58
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$\begingroup$ @KasperAndersen Thank you, that is correct. $\endgroup$– S. Carnahan ♦Commented Jun 30 at 16:01