Difference between revisions of "Gamma'(z)/Gamma(z)=-gamma-1/z+Sum z/(k(z+k))"
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Latest revision as of 05:25, 21 December 2017
Theorem
The following formula holds: $$\dfrac{\Gamma'(z)}{\Gamma(z)} = -\gamma-\dfrac{1}{z}+\displaystyle\sum_{k=1}^{\infty} \dfrac{z}{k(z+k)},$$ where $\Gamma$ denotes gamma, $\gamma$ denotes the Euler-Mascheroni constant, and $\log$ denotes the logarithm.
Proof
References
- 1960: Earl David Rainville: Special Functions ... (previous) ... (next): $9.(1)$
