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How can we show that $f(x)=\Gamma(x+s)/\Gamma(x)$ is an increasing function for $x>1$ and $0<s<1$. I have checked by plotting in sage or wolframalpha. So the result is true for sure. I started proving by showing derivative of $f$ positive for $x>1$. But then we have $$f'(x) = f(x)(\psi(x+s)-\psi(x)),$$ where $\psi$ is digamma function as defined in the literature. But now I need to show $\psi$ is an increasing function, which I again know is true, though i cannot figure out how to show it formally. Any other ways of proving $f$ is an increasing function are welcome.

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$$y=\frac{\Gamma (x+s)}{\Gamma (x)}\implies \log(y)=\log (\Gamma (x+s))-\log (\Gamma (x))$$ Now, using Stirling approximation $$\log (\Gamma (t))=t (\log (t)-1)+\frac{1}{2} \left(\log (2 \pi )-\log \left({t}\right)\right)+\frac{1}{12 t}+O\left(\frac{1}{t^2}\right)$$ apply it twice and continue with Taylor series to get $$\log(y)=s \left(\log(x)-\frac{1}{2 x}-\frac{1}{12 x^2}\right)+O\left(s^2\right)$$ $$y=e^{\log(y)}=1+s \left(\log(x)-\frac{1}{2 x}-\frac{1}{12 x^2}\right)+O\left(s^2\right)$$ and the quantity inside brackets is an increasing function of $x$ and it is positive as soon as $x >1.5$.

Edit

Using Stirling approximation up to $O\left(\frac{1}{t^{21}}\right)$, expanding $\log(y)$ up to $O\left(s^{20}\right)$ and making $s=1$ what is obtained is $$\lim_{s\to 1} \, \log \left(\frac{\Gamma (x+s)}{\Gamma (x)}\right)=\log (x)+\frac{1}{420 x^{20}}+O\left(\frac{1}{x^{21}}\right)$$ that is to say $$\lim_{s\to 1} \, \frac{\Gamma (x+s)}{\Gamma (x)}=x+\frac{1}{420 x^{19}}+O\left(\frac{1}{x^{20}}\right)$$ instead of $x$.

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