View source for Gamma'(z)/Gamma(z)=-gamma-1/z+Sum z/(k(z+k))

==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==
* {{BookReference|Special Functions|1960|Earl David Rainville|prev=Reciprocal gamma written as an infinite product|next=findme}}: $9.(1)$

[[Category:Theorem]]
[[Category:Unproven]]