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

Take the derivative of both side: \begin{eqnarray*} \Gamma' \left({z}\right) & = & -\frac{e^{-\gamma z}(1+\gamma z)}{z^2} \prod_{n \mathop = 1}^\infty \Bigg[ \frac{e^{\frac z n}}{1 + \frac z n} \Bigg] + \frac{e^{-\gamma z}}{z} \sum_{n \mathop=1}^\infty \Bigg[ \frac{z}{n(z+n)} \prod_{i \mathop = 1}^\infty \frac{e^{\frac z i}}{1 + \frac z i} \Bigg] \\ & = &-\frac{e^{-\gamma z}(1 + \gamma z)}{z^2} \Gamma \left({z}\right) \frac{z}{e^{-\gamma z}} + \frac{e^{-\gamma z}}{z} \sum_{n \mathop = 1}^\infty \Bigg[ \frac{z}{n(z+n)} \frac{z}{e^{-\gamma z}} \Gamma \left({z} \right) \Bigg] \\ & = &-\frac{1+\gamma z}{z} \Gamma \left({z} \right) + \sum_{n \mathop = 1}^\infty \Bigg[ \frac{z \Gamma \left({z} \right)}{n(z+n)} \Bigg ] \\ \end{eqnarray*}

Divide both side by $\Gamma \left({z} \right)$: \begin{eqnarray*} \frac{\Gamma' \left({z} \right)}{\Gamma \left({z} \right)} & = & -\frac{1+\gamma z}{z} + \sum_{n \mathop = 1}^\infty \Bigg[ \frac{z}{n(z+n)} \Bigg] \\ & = & -\gamma -\frac 1 {z} + \sum_{n \mathop = 1}^\infty \Bigg[ \frac 1 {n} - \frac 1 {z+n} \Bigg] \\ & = & -\gamma + \sum_{n \mathop = 1}^\infty \Bigg[ \frac 1 {n} - \frac 1 {z+n-1} \Bigg] \end{eqnarray*}