Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 8 down vote accepted

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.

share|improve this answer
1  
This is a nice answer! –  Olivier Oloa Aug 24 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 $$ \log(x)\le\frac{\Gamma'(x+1)}{\Gamma(x+1)}\le\log(x+1) $$

share|improve this answer

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 $$\frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}=\frac{5 (8 x (3 x+1)+1)}{30 x \left(8 x^2+4 x+1\right)+1}+\log (x)$$

share|improve this answer
    
Nice simplification. –  kaka Aug 24 at 8:10
    
Note that all these are asymptotic formulas and non of them proves the inequality for $x>0$. –  Omran Kouba Aug 24 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 at 8:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.