Legendre Duplication by Logarithmic Derivatives

Question

I'm going through Alfor's book for complex analysis. I'm going through and filling in the details for the duplication formula via the logarithmic derivatives. When getting the logarithmic derivative, I'm lost about the $-\frac{1}{z}$ in the equation. $$\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod\limits_{k=1}^\infty \frac{k}{k+z} e^\frac{z}{k}$$ $$\Gamma'(z) = (-\frac{1}{z} - \gamma + \sum\limits_{n=1}^\infty \frac{1}{n} - \frac{1}{n+z}) \Gamma(z)$$ $$\frac{d}{dz} \log \Gamma(z) = -\frac{1}{z} - \gamma + \sum\limits_{n=1}^\infty \frac{z}{n(n+z)}$$ Since we want to look at the second derivative, $$\frac{d^2}{dz^2} \log \Gamma(z) = \frac{1}{z^2} + \sum\limits_{n=1}^\infty \frac{1}{{(n+z)}^2} \neq \sum \frac{1}{{(n+z)}^2}$$ I was trying to find other info and found the same property for $\psi(z + 1) = \frac{1}{z} + \psi(z)$ where $\psi$ is the digamma function ($\psi(z) = \frac{d}{dz} \log \Gamma(z)$). Is Alfor's supposed to be considering $\log \Gamma(z+1)$ that is being twice differentiated? This is an issue even if I try summing the $z$ and $z+\frac{1}{2}$ terms, because there is a $\frac{1}{z^2} + \frac{1}{(z+\frac{1}{2})^2}$ that doesn't match with the $\frac{1}{4z^2}$ from the second derivative of $\log \Gamma(2z)$.

Answer 1

The issue was I did not write out the indices on the summations. Alfors has the sums going from 0 to $\infty$ because he scooted the residual $\frac{1}{z^2}$ term inside since it is adding 0 to it. This means the second derivative of $\log \Gamma(z)$ is indeed $\sum\limits_{n=0}^\infty \frac{1}{{(z+n)}^2}$