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I am looking for an upper bound for

$$ \frac{\Gamma(x+\beta)}{\Gamma(x)},\,\,\,\beta>0.$$

In this question it was shown that

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

Then, I believe that there must be some sort of polynomial upper bound but I have failed to come up with one. This is true for the case when $\beta$ is an integer. Any suggestion would be appreciated.

Community
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Gabo
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3 Answers3

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In fact, $$ \Gamma(x+\beta)\le x^\beta \Gamma(x). $$ See a proof in a short note published by James G. Wendel in 1948.

J. G. Wendel (1948): Note on the Gamma function, Amer. Math. Monthly. Vol. 55, No. 9, pp. 563--564.

DOI: 10.2307/2304460.

https://www.jstor.org/stable/2304460

Egor Maximenko
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If $\beta$ is an integer then from $\Gamma(z+1) = z\Gamma(z)$ we have

$$ \frac{\Gamma(x+\beta)}{\Gamma(x)} = \prod_{k=0}^{\beta-1} (x+k). $$

Otherwise we can use the monotonicity of $\Gamma$ can obtain the crude bound

$$ \frac{\Gamma(x+\beta)}{\Gamma(x)} = \frac{\Gamma(x + \beta - \lceil\beta\rceil)}{\Gamma(x)} \prod_{k=1}^{\lceil\beta\rceil} (x+\beta-k) \leq \prod_{k=1}^{\lceil\beta\rceil} (x+\beta-k) $$

which holds for $x + \beta - \lceil\beta\rceil \geq x_0$, where $x_0 \approx 1.46163$ is the location of the minimum of the gamma function on $\mathbb{R}^+$.

Antonio Vargas
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  • For my specific purpose this answer is enough, then I am going to accept it. Thanks. – Gabo Jul 17 '13 at 13:55
  • @Gabo You might want to wait a day or so before accepting an answer. It's possible that someone who can give a better estimate than mine just hasn't seen the question yet! :) – Antonio Vargas Jul 17 '13 at 14:07
  • In particular it looks like $\Gamma(x+\beta) \leq x^\beta$ for all $x \geq 0$ if $0 < \beta < 1$, which could be used to improve my bound quite a bit. I don't see how to prove it though. – Antonio Vargas Jul 17 '13 at 14:13
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Note that $$\Gamma(x + 1) \sim {x^x e^{-x}\over \sqrt{2\pi x}}$$ by the Stirling formula. So $${\Gamma(x + 1 + \beta )\over \Gamma(x + 1)}\sim {(x + \beta)^{x + \beta}\over x^x}\cdot{\sqrt{2\pi x}\over\sqrt{2\pi (x + \beta)}}\cdot e^{-\beta} \sim (1 + \beta/x)^x e^{-\beta} \cdot (x + \beta)^\beta \sim(x + \beta)^\beta \sim x^\beta.$$

ncmathsadist
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