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:

 $\displaystyle \operatorname{Log} \left(\Gamma \left({z}\right)\right)$ $=$ $\displaystyle -\gamma \left(z-1\right) + \sum_{k \mathop = 2}^\infty \frac{\left({-1}\right)^k \zeta \left({k}\right)}{k} \left(z-1\right)^k$

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

Proof

$\Gamma\left(z+1\right)=z\Gamma\left(z\right)$

Hence:

 $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z+1}\right)\right)$ $=$ $\displaystyle \frac {\d} {\d z}\operatorname{Log} \left(z\Gamma \left({z}\right)\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\d} {\d z}\operatorname{Log} \left(\Gamma \left({z}\right)\right) + \operatorname{Log}\left(z\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z}\right)\right) + \frac {\d} {\d z} \operatorname{Log}\left(z\right)$ Linearity of the Derivative $\displaystyle$ $=$ $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z}\right)\right) + \frac 1 z$ Derivative of Natural Logarithm Function
Succesive use of this identity gives us
 $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z}\right)\right)$ $=$ $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z+M}\right)\right) - \sum_{k \mathop = 0}^{M-1} \frac{ 1} {z+k}$
And thus from the Linearity of the derivative gives us
 $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({z}\right)\right)$ $=$ $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({z+M}\right)\right) + \sum_{k \mathop = 0}^{M-1} \frac {\left(-1\right)^{n}\left(n-1\right)!}{\left(z+k\right)^n}$
 $\displaystyle \operatorname{Log} \left(\Gamma \left({z + 1}\right)\right)$ $=$ $\displaystyle z \operatorname{Log} \left(z\right) - z + \frac 1 2 \left(\operatorname{Log} \left(z\right)\right) + O \left(\frac 1 z\right)$ $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z + 1}\right)\right)$ $=$ $\displaystyle \operatorname{Log} \left({z}\right) + O \left({\frac 1 z}\right)$ $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({z + 1}\right)\right)$ $=$ $\displaystyle O \left({\frac 1 z}\right)$ $\displaystyle \frac {\d} {\d z} \operatorname{Log} \left(\Gamma \left({z + M}\right)\right) - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$ $=$ $\displaystyle \operatorname{Log}\left({z + M}\right) - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + O \left({\frac 1 {z + M} }\right)$ $\displaystyle \lim_{M \to \infty} \frac {\d} {\d z} \operatorname{Log} \left({\Gamma \left({z + M}\right)}\right) - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$ $=$ $\displaystyle \lim_{M \to \infty} \operatorname{Log}\left({z + M}\right) - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + lim_{M \to \infty} O \left({\frac 1 {z + M} }\right)$ $\displaystyle$ $=$ $\displaystyle \lim_{M \to \infty} \operatorname{Log}\left({z + M}\right) - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$

which by definition of the constant gives us:

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

Also:

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

shows that:

$\lim_{M \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \operatorname{Log} \left(\Gamma \left({M+1}\right)\right)= 0$

thus for $n > 1$:

 $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({1}\right)\right)$ $=$ $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({1+M}\right)\right) + \sum_{k \mathop = 1}^{M} \frac {\left(-1\right)^{n}\left(n-1\right)!}{\left(k\right)^n}$ $\displaystyle \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({1}\right)\right)$ $=$ $\displaystyle \lim_{M \to \infty} \frac {\d^n} {\d z^n} \operatorname{Log} \left(\Gamma \left({1+M}\right)\right) + \sum_{k \mathop = 1}^M \frac {\left(-1\right)^{n}\left(n-1\right)!}{\left(k\right)^n}$ $\displaystyle$ $=$ $\displaystyle \zeta\left(n\right)\left(n-1\right)!(-1)^n$

Thus by definition of Taylor series:

 $\displaystyle \operatorname{Log} \left(\Gamma \left({z}\right)\right)$ $=$ $\displaystyle \operatorname{Log} \left(\Gamma \left({1}\right)\right)-\gamma \left(z-1\right) + \sum_{k \mathop = 2}^\infty \frac{\left({-1}\right)^k \zeta \left({k}\right)\left(k-1\right)!}{k!} \left(z-1\right)^k$ $\displaystyle$ $=$ $\displaystyle -\gamma \left(z-1\right) + \sum_{k \mathop = 2}^\infty \frac{\left({-1}\right)^k \zeta \left({k}\right)}{k} \left(z-1\right)^k$

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.

$\blacksquare$