How may one go about proving


(away from the poles) directly? By a direct proof, I mean not to go through the usual Stirling formula with the exact error term. The use of a rough form of Stirling's formula is welcome.


You can use the product representation of $\Gamma(z)$, take logs and differentiate the resulting series.

$$\Gamma(z) = \dfrac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty} \left(1 + \dfrac{z}{n}\right)^{-1} \ e^{\frac{z}{n}}$$

You can find more information here: Polygamma Function and Digamma Function.

  • $\begingroup$ Thanks for the answer. From the product formula I get $\frac{\Gamma'(z)}{\Gamma(z)}=-\gamma-\frac1z+\sum_{n\geq1}\frac{z}{n(z+n)}$ and I don't know where to go from here. Is there a similar partial fractions series for the logarithm so that one can compare the two? $\endgroup$ – timur Dec 24 '10 at 22:33
  • $\begingroup$ @Timur: The series is very closely related to Harmonic numbers, which are usually compared to the integral of $1/z$. $\endgroup$ – Aryabhata Dec 25 '10 at 1:44
  • $\begingroup$ I am sure I am being stupid but the harmonic numbers are indexed by integers, while here we have a complex argument in the digamma (btw thanks for the link, I did not know this function is called digamma). Collapsing everything onto the real line would give something in terms of the real parts but the inequality we what to prove involve absolute value.. $\endgroup$ – timur Dec 25 '10 at 3:08
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    $\begingroup$ @timur: You should be able bound ∣∣1/(n+z)∣∣ in terms of 1/(n+∣∣z∣∣) and 1/(n−∣∣z∣∣) for sufficiently large n. I confess, I haven't gone deeper into the details, though, but I am quite confident that this approach will work. $\endgroup$ – Aryabhata Dec 25 '10 at 3:36
  • $\begingroup$ @timur: the harmonic numbers and the digamma function are essentially the same thing, the only difference is the Euler-Mascheroni constant. $\endgroup$ – J. M. is a poor mathematician Dec 25 '10 at 4:45

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