Upper bound for the quotient of gamma functions?

Question

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.

Answer 1

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 $\mathbb{R}^+$.

Answer 2

Note that $$\Gamma(x + 1) \sim {x^x e^{-x}\over \sqrt{2\pi x}}$$ by the Stirling formula. So $${\Gamma(x + 1 + \beta )\over \Gamma(x + 1)}\sim {(x + \beta)^{x + \beta}\over x^x}\cdot{\sqrt{2\pi x}\over\sqrt{2\pi (x + \beta)}}\cdot e^{-\beta} \sim (1 + \beta/x)^x e^{-\beta} \cdot (x + \beta)^\beta \sim(x + \beta)^\beta \sim x^\beta.$$