Derivative of Gamma Function at 1
Theorem
For the Gamma function $\Gamma$ and the Euler-Mascheroni constant $\gamma$, the following holds: $$ \Gamma ' (1) = - \gamma $$
Proof 1
The derivative of the Gamma function times its reciprocal: $$ {{ \Gamma ' (z) } \over { \Gamma (z) }} = - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right) $$
Substituting $z = 1$ into the product of the derivative and the reciprocal of the Gamma function gives $$ \begin{align*} {{ \Gamma ' (1) } \over { \Gamma (1) }} =& - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { 1 + n - 1 }} \right) \\ =& - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { n }} \right) \\ =& - \gamma + 0 \end{align*} $$ and since $\Gamma (1) = 0! = 1$, we obtain $\Gamma ' (1) = - \gamma$.
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