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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.2010-10-27
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    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.2010-10-27

1 Answers 1

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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 $\gamma=\lim_{n\to\infty}\sum_{j=1}^n \frac1{j}-\log n$.2010-10-27
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    M: Oh ho! I missed the killer blow!2010-10-28