Derivative of Gamma Function at 1

Theorem

Let $\Gamma$ denote the Gamma function.

Then:

$\map {\Gamma'} 1 = -\gamma$

where:

$\map {\Gamma'} 1$ denotes the derivative of the Gamma function evaluated at $1$

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

Setting $n = 1$:

$\ds \frac {\map {\Gamma'} 1} {\map \Gamma 1}$

$=$

$\ds -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {1 + n - 1} }$

$\ds $

$=$

$\ds -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 n}$

$\ds $

$=$

$\ds -\gamma + 0$

$\ds $

$=$

$\ds -\gamma$

$\map \Gamma 1 = \paren {1 - 1}! = 1$

Hence:

$\map {\Gamma'} 1 = -\gamma \map \Gamma 1 = -\gamma$

$\blacksquare$