From Gamma Difference Equation: :$\map \Gamma {z + 1} = z \map \Gamma z$ Hence: {{begin-eqn}} {{eqn | l = \frac {\d} {\d z} \map \Log {\map \Gamma {z + 1} }       | r = \frac {\d} {\d z} \map \Log {z \map \Gamma z}       | c =  }} {{eqn | r = \frac {\d} {\d z} \map \Log {\map \Gamma z} + \map \Log z       | c =  }} {{eqn | r = \frac {\d} {\d z} \map \Log {\map \Gamma z} + \frac {\d} {\d z} \map \Log z       | c = Sum Rule for Derivatives }} {{eqn | r = \frac {\d} {\d z} \map \Log {\map \Gamma z} + \frac 1 z       | c = Derivative of Natural Logarithm Function }} {{end-eqn}} Successive use of this identity gives us {{begin-eqn}} {{eqn | l = \frac {\d} {\d z} \map \Log {\map \Gamma z}       | r = \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}       | c =  }} {{end-eqn}} and thus from the Sum Rule for Derivatives: {{begin-eqn}} {{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma z}       | r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + M} } + \sum_{k \mathop = 0}^{M - 1} \frac {\paren {-1}^n \paren {n - 1}!} {\paren {z + k}^n} }} {{end-eqn}} From Stirling's Formula for Gamma Function: {{begin-eqn}} {{eqn | l = \map \Log {\map \Gamma {z + 1} }       | r = z \map \Log {\map \Gamma z} - z + \frac 1 2 \map \Log z + \map \OO {\frac 1 z}       | c =  }} {{eqn | l = \frac {\d} {\d z} \map \Log {\map \Gamma {z + 1} }       | r = \map \Log z + \map \OO {\frac 1 z}       | c =  }} {{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + 1} }       | r = \map \OO {\frac 1 z}       | c =  }} {{eqn | l = \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}       | r = \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \map \OO {\frac 1 {z + M} }       | c =  }} {{eqn | l = \lim_{M \to \infty} \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}       | r = \lim_{M \to \infty} \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \lim_{M \to \infty} \map \OO {\frac 1 {z + M} } }} {{eqn | r = \lim_{M \to \infty} \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }} {{end-eqn}} which by definition of the Euler-Mascheroni constant: :$\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma 1} = \lim_{M \to \infty} \frac {\d} {\d z} \map \Log {\map \Gamma {1 + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k} = -\gamma$ Also: :$\dfrac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$ shows that: :$\displaystyle \lim_{M \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$ thus for $n > 1$: {{begin-eqn}} {{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}       | r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}       | c =  }} {{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}       | r = \lim_{M \to \infty} \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n} }} {{eqn | r = \map \zeta n \paren {n - 1}! \paren {-1}^n       | c =  }} {{end-eqn}} Thus by definition of Taylor series: {{begin-eqn}} {{eqn | l = \map \Log {\map \Gamma z}       | r = \map \Log {\map \Gamma 1} - \gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k \paren {k - 1}!} {k!} \paren {z - 1}^k  }} {{eqn | r = -\gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac{\paren {-1}^k \map \zeta k} k \paren {z - 1}^k }} {{end-eqn}} 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 $\cmod {1 - 0} = 1$. {{qed}} Category:Gamma Function Category:Natural Logarithms bw2a974j6jwrkswpkzjbml5mgy5zseg