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.

References

• 1960: Earl David Rainville: Special Functions ... (previous) ... (next): $9.(1)$