Reciprocal times Derivative of Gamma Function

Theorem
Let $\Gamma$ denote the Gamma function.

Then:


 * $\displaystyle \dfrac {\Gamma\,' \left({z}\right)} {\Gamma \left({z}\right)} = -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 {z + n - 1} }\right)$

where:
 * $\Gamma\,' \left({z}\right)$ denotes the derivative of the Gamma function
 * $\gamma$ denotes the Euler-Mascheroni constant.

Proof
From the Weierstrass form of the Gamma function:
 * $\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left({\left({1 + \frac z n}\right) e^{-z / n} }\right)$

Taking the reciprocal of both sides:
 * $\displaystyle \Gamma \left({z}\right) = \frac {e^{-\gamma z}} z \prod_{n \mathop = 1}^\infty \frac {e^{z/n}} {1 + \frac z n}$

Taking the derivative of both sides:

Dividing both sides by $\Gamma \left({z}\right)$: