Approximation to Reciprocal times Derivative of Gamma Function

Theorem
Let $\Gamma$ denote the gamma function.

For all $z \in \C$ such that $\left\vert{\arg\left({z}\right)}\right\vert < \pi - \epsilon, \left\vert{z}\right\vert > 1$:


 * $\dfrac {\Gamma' \left({z}\right)} {\Gamma \left({z}\right)} = \ln z + \mathcal O \left({z^{-1}}\right)$

where:
 * $\mathcal O \left({z^{-1}}\right)$ denotes big-O notation
 * the implied constant depends on $\epsilon$.

Proof
From Stirling's Formula for Gamma Function:


 * $\ln \Gamma(z) = \left({z - \dfrac 1 2}\right) \ln z - z + \dfrac {\ln 2 \pi} 2 + \mathcal O \left({z^{-1}}\right)$

Taking the derivative:


 * $(1): \quad \dfrac{\Gamma' \left({z}\right)} {\Gamma \left({z}\right)} = \ln z - \dfrac 1 {2 z} + \dfrac {\mathrm d} {\mathrm dz} \mathcal O \left({z^{-1}}\right)$

Since there is $c \left({\epsilon}\right) > 0$ such that:


 * $- \dfrac c {\left\vert{z^{-1} }\right\vert} < \left\vert{\mathcal O \left({z^{-1} }\right)}\right\vert < \dfrac c{\left\vert{z^{-1} }\right\vert}, \quad \left\vert{z}\right\vert > 1$

it follows directly that the third term in $(1)$ is $\mathcal O \left({z^{-1}}\right)$.