# An uncanny inequality with Gamma function

Prove for $x>0$ that $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}>\log x$$

How to prove this inequality? thanks. This is a problem from Miklos Schweitzer Competition.

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Let $~g(x)=\dfrac{\Gamma'(x+1)}{\Gamma(x+1)}-\log x$. Now we have \eqalign{g'(x)&=-\frac{1}{x}+\sum_{k=1}^\infty\frac{1}{(x+k)^2}\cr &<-\frac{1}{x}+\sum_{k=1}^\infty\int_{k-1}^k\frac{dt}{(x+t)^2}\cr &=-\frac{1}{x}+\int_0^\infty\frac{dt}{(x+t)^2}=0 } So $g$ is decreasing on $(0,+\infty)$. Moreover $$g(n)=\sum_{k=1}^n\frac{1}{k}-\gamma-\log n$$ So $\lim_{n\to\infty}g(n)=0$ and consequently $\lim_{x\to\infty}g(x)=0$,and because $g$ is decreasing on $(0,+\infty)$ we conclude that $g >0$ on this interval and we are done.

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This is a nice answer! – Olivier Oloa Aug 24 '14 at 8:35

Since $\log(\Gamma(x))$ is convex, we have $$\frac{\log(\Gamma(x+1))-\log(\Gamma(x))}{(x+1)-x}\le\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x+1))\le\frac{\log(\Gamma(x+2))-\log(\Gamma(x+1))}{(x+2)-(x+1)}$$ which can be rewritten, using $\Gamma(x+1)=x\Gamma(x)$, as $$\log(x)\le\frac{\Gamma'(x+1)}{\Gamma(x+1)}\le\log(x+1)$$

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If you use Stirling approximation for $n!$, you then have $$\Gamma(1+x)\simeq \sqrt{2 \pi } e^{-x} x^{x+\frac{1}{2}}$$ Computing the derivative $$\Gamma'(1+x)\simeq \sqrt{\frac{\pi }{2}} e^{-x} x^{x-\frac{1}{2}} (2 x \log (x)+1)$$ so, after simplification, $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}=\frac{1}{2 x}+\log (x)$$ If, instead, you use Gosper approximation $$\Gamma(1+x)\simeq \sqrt{\pi } e^{-x} x^x \sqrt{2 x+\frac{1}{3}}$$ you should arrive to $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}=\frac{3}{6 x+1}+\log (x)$$ If you use Burnside approximation $$\Gamma(1+x)\simeq \sqrt{2 \pi } e^{-x-\frac{1}{2}} \left(x+\frac{1}{2}\right)^{x+\frac{1}{2}}$$ you should arrive to $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}=\log \left(x+\frac{1}{2}\right)$$ If you use Ramanujan approximation $$\Gamma(1+x)\simeq \sqrt{\pi } e^{-x} x^x \sqrt[6]{8 x^3+4 x^2+x+\frac{1}{30}}$$ you should arrive to $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}=\frac{5 (8 x (3 x+1)+1)}{30 x \left(8 x^2+4 x+1\right)+1}+\log (x)$$

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Nice simplification. – kaka Aug 24 '14 at 8:10
Note that all these are asymptotic formulas and non of them proves the inequality for $x>0$. – Omran Kouba Aug 24 '14 at 8:32
@OmranKouba. I agree partly with you since they are very good approximations. I thought that it was simpler than to prove that $\psi ^{(0)}(x+1) >\log (x)$ – Claude Leibovici Aug 24 '14 at 8:36