# 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 $1$
$\gamma$ denotes the Euler-Mascheroni constant.

## Proof 1

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

Setting $n = 1$:

 $\displaystyle \frac {\Gamma' \left({1}\right)} {\Gamma \left({1}\right)}$ $=$ $\displaystyle -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 {1 + n - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle -\gamma + \sum_{n \mathop = 1}^\infty \left({\frac 1 n - \frac 1 n}\right)$ $\displaystyle$ $=$ $\displaystyle -\gamma + 0$ $\displaystyle$ $=$ $\displaystyle -\gamma$

$\blacksquare$