Derivative of Gamma Function at 1/Proof 1

Theorem
Let $\Gamma$ denote the Gamma function.

Then:


 * $\Gamma\,' \left({1}\right) = -\gamma$

where:
 * $\Gamma\,' \left({1}\right)$ denotes the derivative of the Gamma function evaluated at $1$
 * $\gamma$ denotes the Euler-Mascheroni constant.

Proof
From Reciprocal times Derivative of Gamma Function:


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

Setting $n = 1$: