Taylor Series of Logarithm of Gamma Function

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

Let $\zeta \left({s}\right)$ denote the Riemann zeta function.

Let $\Gamma \left({z}\right)$ denote the gamma function.

Let $\operatorname{Log}$ denote the natural logarithm.

Then $\operatorname{Log} \left({\Gamma \left({z}\right)}\right)$ has the power series expansion:

valid for all $z \in \C$ such that $\left\lvert{z - 1}\right\rvert < 1$.

Proof
From Gamma Difference Equation:
 * $\Gamma\left(z+1\right)=z\Gamma\left(z\right)$

Hence:


 * Succesive use of this identity gives us


 * And thus from the Linearity of the derivative gives us

From Stirling's Formula for Gamma Function:

which by definition of the constant gives us:


 * $\displaystyle \frac {\rd} {\rd z} \operatorname{Log} \left(\Gamma \left({1}\right)\right)= \lim_{M \to \infty} \frac {\rd} {\rd z} \operatorname{Log} \left(\Gamma \left({1+M}\right)\right) - \sum_{k \mathop = 0}^{M-1} \frac{ 1} {1+k} = -\gamma$

Also:


 * $\frac {\rd^{1+k}} {\rd z^{1+k}} \operatorname{Log} \left(\Gamma \left({z+1}\right)\right)= O\left(\frac 1 z\right)$

shows that:
 * $\lim_{M \to \infty} \frac {\rd^{1+k}} {\rd z^{1+k}} \operatorname{Log} \left(\Gamma \left({M+1}\right)\right)= 0$

thus for $n > 1$:

Thus by definition of Taylor series:

From Zeroes of Gamma Function, we see the Gamma function is everywhere non-zero.

Thus $\operatorname{Log} \left(\Gamma \left({z}\right)\right)$ has poles only where Gamma does, that is, the nonpositive integers

Thus 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 with have that its radius of convergence is |1-0|=1.