# Stirling-type formula for the logarithmic derivative of the Gamma function

How may one go about proving

$\displaystyle\frac{\Gamma'(s)}{\Gamma(s)}=O(\log|s|)$,

(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.

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

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

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? –  timur Dec 24 '10 at 22:33
@Timur: The series is very closely related to Harmonic numbers, which are usually compared to the integral of $1/z$. –  Aryabhata Dec 25 '10 at 1:44