Upper bound for the quotient of gamma functions? 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.
 A: 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 $\mathbb{R}^+$.
A: Note that 
$$\Gamma(x + 1) \sim {x^x e^{-x}\over \sqrt{2\pi x}}$$
by the Stirling formula.  So
$${\Gamma(x + 1 + \beta )\over \Gamma(x + 1)}\sim {(x + \beta)^{x + \beta}\over
x^x}\cdot{\sqrt{2\pi x}\over\sqrt{2\pi (x + \beta)}}\cdot e^{-\beta}
\sim (1 + \beta/x)^x e^{-\beta} \cdot (x + \beta)^\beta
\sim(x + \beta)^\beta \sim x^\beta.$$
A: 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
