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I know I have asked a similar question a couple of days, ago, but I still have a problem. I need a upper bound for: $$ \ln\frac{\Gamma\left(\frac{x}{3}+1\right)}{\Gamma\left(\frac{x}{4}\right)\Gamma\left(\frac{x}{12}\right)} $$ Similar to this bound: $$ \ln\left(\frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}\right)\leq \left(\frac{\ln 4}{3}-\frac{\ln 3}{4}\right)x $$ I tried - as I was adviced in my previous post - applying Stirling's approximation ($ \Gamma(x)\approx\sqrt{\frac{2 \pi}{x}}{\left( \frac{x}{e} \right)}^x $) and using the fact that $ \ln\Gamma(x) $ is a convex function and the Bohr-Mollerup theorem. After much computations, I derived: $$ \ln\frac{\Gamma\left(\frac{x}{3}+1\right)}{\Gamma\left(\frac{x}{4}\right)\Gamma\left(\frac{x}{12}\right)}\le\ln{x}+x\left(\frac{\ln{4}}{4}-\frac{\ln{3}}{3}+\frac{\ln{12}}{12}\right) $$ But I made error somewhere (and I am not able to find it), because this inequality fails for example for $ x=4567 $... So could someone compute the upper bound for me and maybe trace when I could make the mistake? I really need these bounds. Thank you in advance... Hope someone will help me!

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    $\begingroup$ Given the answer to your previous question, you could use $\Gamma(t+1)=t \Gamma(t)$ $\endgroup$ – Henry Aug 22 '15 at 23:02
  • $\begingroup$ Yes, I USED IT, but it yielded the result I have written. And you should read my question more carefully, as I have written also, that I used the Bohr-Mollerup theorem, which is exactly $ \Gamma(t+1)=t\Gamma(t) $. $\endgroup$ – grzegorzmarciakowski Aug 23 '15 at 7:44
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We have the estimates $$ 0\le\log(\Gamma(x))-\left[x\log(x)-x+\frac12\log(2\pi)-\frac12\log(x)\right]\le\frac1{12x} $$ and $$ -\frac1{360x^3}\le\log(\Gamma(x))-\left[x\log(x)-x+\frac12\log(2\pi)-\frac12\log(x)+\frac1{12x}\right]\le0 $$ Using these, along with the fact that $\log\left(\frac{\Gamma\left(\frac x3+1\right)}{\Gamma\left(\frac x4\right)\Gamma\left(\frac x{12}\right)}\right)\sim\log\left(\frac{x^2}{48}\right)$ for $x$ near $0$, $$ \begin{align} \log\left(\frac{\Gamma\left(\frac x3+1\right)}{\Gamma\left(\frac x4\right)\Gamma\left(\frac x{12}\right)}\right) &\le\left(\frac14\log(4)+\frac1{12}\log(12)-\frac13\log(3)\right)x+\frac32\log(x)-\log(12\sqrt{2\pi}) \end{align} $$

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    $\begingroup$ Thank you so much, you saved my thesis! $\endgroup$ – grzegorzmarciakowski Aug 23 '15 at 10:33

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