# ratio of gamma function

How can we show that $$f(x)=\Gamma(x+s)/\Gamma(x)$$ is an increasing function for $$x>1$$ and $$0. I have checked by plotting in sage or wolframalpha. So the result is true for sure. I started proving by showing derivative of $$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.

$$y=\frac{\Gamma (x+s)}{\Gamma (x)}\implies \log(y)=\log (\Gamma (x+s))-\log (\Gamma (x))$$ 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 $$x >1.5$$.
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 $$x$$.