Taylor Series of Logarithm of Gamma Function

Theorem
Let $\gamma$ denote the Euler-Mascheroni constant.

Let $\map \zeta s$ denote the Riemann zeta function.

Let $\map \Gamma z$ denote the gamma function.

Let $\Log$ denote the natural logarithm.

Then $\map \Log {\map \Gamma z}$ has the power series expansion:

which is valid for all $z \in \C$ such that $\cmod {z - 1} < 1$.

Proof
Hence:

Therefore:

Setting:
 * $z = \paren {x - \floor x} \leadsto \d z = \d x$
 * $M = \floor x + 1$:

we have:
 * $z + M = x + 1$

Hence:

Recall Stirling's Formula for Gamma Function:

We now have:
 * $\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} } = \map \Log z + \map \OO {\frac 1 z}$

Therefore:

We now have:
 * $\ds \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {z + M} } = \lim_{M \mathop \to \infty} \map \Log {z + M} $

Setting $z = 1$ into $(2)$:

Also:
 * $\ds \frac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$

shows that:
 * $\ds \lim_{M \mathop \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$

thus for $n > 1$:

Thus by definition of Taylor series:

From Zeroes of Gamma Function, we see that $\map \Gamma z$ is non-zero everywhere.

Thus $\map \Log {\map \Gamma z}$ has poles only where $\Gamma$ does, that is, the negative integers.

Since the radius of convergence of a power series is equal to the distance of its center to the closest point where the function is not analytic:

The radius of convergence of $\map \Log {\map \Gamma z}$ is $\cmod {1 - 0} = 1$.