Reciprocal times Derivative of Gamma Function

Theorem
Let $\Gamma$ denote the Gamma function.

Then:


 * $\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$

where:
 * $\map {\Gamma'} z$ denotes the derivative of the Gamma function
 * $\gamma$ denotes the Euler-Mascheroni constant.

Proof
From the Weierstrass form of the Gamma function:
 * $\ds \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$

Taking the reciprocal of both sides:
 * $\ds \map \Gamma z = \frac {e^{-\gamma z}} z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n}$

Taking the derivative of both sides:

Dividing both sides by $\map \Gamma z$: