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For each $n>0$, how do we prove that $$\Gamma'(n+1)> \log{n} \cdot \Gamma(n+1)$$

  • I had spent about half an hour on this question, but just could find any way of proceeding for the solution.

  • Wikipedia page gave me an interesting identity $$\Gamma'(n+1)= n! \cdot \Biggl( - \gamma + \sum\limits_{k=1}^{n} \frac{1}{k}\Biggr)$$ But i don't know how it can be applied here.

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A link to the wikipedia page would be handy. – AD. Oct 27 '10 at 19:52
If the identity holds we would only need to prove that $\log n <\sum \frac{1}{k} -\gamma$, i.e. $$\sum \frac{1}{k}-\log n +\phi(n)>\gamma$$ for some positive function $\phi$. Finding such a $\phi$ might be done using some refinement in the estimate of the integral test, I believe. – AD. Oct 27 '10 at 19:58
up vote 6 down vote accepted

Well, if we rearrange the inequality to

$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)} > \log n$$

and then note that

$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{j=1}^n \frac1{j}$$

you should be able to easily deduce the inequality.

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Note also that $\gamma=\lim_{n\to\infty}\sum_{j=1}^n \frac1{j}-\log n$. – J. M. Oct 27 '10 at 20:16
M: Oh ho! I missed the killer blow! – anonymous Oct 28 '10 at 5:09

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