# Upper bound of the fraction of gamma functions

Is there a simple upper bound of the following fraction of gamma functions for any $$a,b\geq1/2$$: $$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}$$ An upper bound in the following form is enough: there exists a constant $$C>0$$ and a function $$f(a,b)$$ such that: $$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\leq C\cdot f(a,b), \forall a,b\geq1/2$$

I have a guess of $$f$$, which is $$\frac{a+b}{a}$$. Is it true?

$$\newcommand{\Ga}{\Gamma}$$
The inequality in question trivially holds if e.g. $$C=1$$ and $$f(a,b)$$ equals the left-hand side of the inequality.
More informatively, the bound with $$f(a,b)=\frac{a+b}a$$ indeed holds: $$$$\frac{\Ga(a+b)}{a\Ga(a)\Ga(b)}\le C^a\Big(\frac{a+b}a\Big)^a$$$$ for some real $$C>0$$ and all $$a,b\ge1/2$$. Indeed, by Stirling's formula for the gamma function, $$$$\Ga(x)\asymp\frac1{\sqrt x}\,\Big(\frac xe\Big)^x$$$$ for $$x\ge1/2$$, where $$\asymp$$ means "equals up to a universal positive constant factor". So, $$\begin{multline} \frac{\Ga(a+b)}{a\Ga(a)\Ga(b)}\asymp\frac1a\,\sqrt{\frac{ab}{a+b}}\frac{(a+b)^{a+b}}{a^a b^b} \ll \Big(\frac{a+b}a\Big)^a \Big(\frac{a+b}b\Big)^b \\ = \Big(\frac{a+b}a\Big)^a \Big(1+\frac ab\Big)^b < e^a\Big(\frac{a+b}a\Big)^a \end{multline}$$ for $$a,b\ge1/2$$.
• Shouldn't $\Gamma(x)\asymp\sqrt{x}(x/e)^x$? – neverevernever Oct 18 '18 at 16:21
• @neverevernever : No, it is as I wrote. You may be confusing it with the Stirling formula for $n!=\Gamma(n+1)$. I have added the reference to Stirling's formula; see the last displayed formula in Section "Stirling's formula for the gamma function" there. – Iosif Pinelis Oct 18 '18 at 21:10