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 Weierstrass Form:
 * $\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left[{\left({1 + \frac z n}\right) e^{-z / n} }\right]$

We can take the reciprocal of the both side and obtain:
 * $\displaystyle \Gamma \left({z}\right) = \frac {e^{-\gamma z}}{z} \prod_{n \mathop = 1}^\infty \frac{e^{z/n}}{1 + \frac{z}{n} }$

Take the derivative of both side:

Divide both side by $\Gamma \left({z} \right)$: