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I'm sorry if this is a simple question, but this page on Wolfram Research states that it follows from Stirling's formula that:

$$ \frac{\Gamma(x+\beta)}{\Gamma(x)} \approx x^\beta $$

for large $x$, but I'm just not seeing a simple derivation. Could you help me see how this follows (other resources I've seen state that the more general formulation provided on that link, that

$$\frac{\Gamma(x+\beta)}{\Gamma(x+\alpha)} \approx x^{\beta-\alpha}$$

follows from some algebra using Stirling's formula, but again, I can't come up with it.

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I don't know if you've seen this paper, but in "The asymptotic expansion of a ratio of Gamma functions", they get a similar expansion than the one you're looking for, albeit without using Stirling's formula. You can find the paper at projecteuclid.org/DPubS/Repository/1.0/… –  M Turgeon Mar 8 '12 at 18:24
The more general formulation follows almost trivially from the first, since $\frac{\Gamma(x+\beta)}{\Gamma(x+\alpha)} = \frac{\Gamma(x+\beta)}{\Gamma(x)}\cdot\frac{\Gamma(x)}{\Gamma(x+\alpha)}$ –  Steven Stadnicki Mar 8 '12 at 20:52

1 Answer 1

up vote 8 down vote accepted

We will write $\gamma=\beta-\alpha$ and plug in Stirling's:

$$\frac{\Gamma(x+1+\beta)}{\Gamma(x+1+\alpha)} \approx \frac{\sqrt{2\pi(x+\beta)} \left(\frac{x+\beta}{e}\right)^{ x+\beta} }{\sqrt{2\pi (x+\alpha)}\left(\frac{x+\alpha}{e}\right)^{x+\alpha}} $$

$$ =\left(1+\frac{\gamma}{x+\alpha}\right)^{ x+\alpha+1/2} (1+\beta/x)^{\gamma}\left(\frac{x}{e}\right)^{\gamma} \approx e^\gamma (x/e)^\gamma = x^\gamma. $$

Above we regrouped terms and used $e^u\approx (1+u/n)^n$ and $1+u/n\approx 1$ for large $n$.

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Another route is induction on Gautschi's inequality. –  anon Mar 8 '12 at 20:46

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