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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?

flag	asked Mar 12 2011 at 4:54 7-adic 1,107●5●16

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I may be missing something, but how about en.wikipedia.org/wiki/… – Yemon Choi Mar 12 2011 at 5:08

S. Carnahan · Accepted Answer · 2011-03-12 05:48:38Z UTC

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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:

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

answered Mar 12 2011 at 5:48

S. Carnahan♦
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I would say that equivalent was obtained through the generalized Stirling formula for the $\Gamma$ function, as pointed to by Yemon Choi. – Julien Puydt Mar 12 2011 at 15:20

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That seems to be a reasonably likely explanation. I just Googled for Pochhammer asymptotic, and didn't bother to think too much. – S. Carnahan♦ Mar 12 2011 at 15:43

What is the Stirling formula for x(x+1)(x+2)…(x+n-1)?

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