partial fraction series for digamma function

Proof: Start with$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k},$$ so$$ \ln\Gamma(z)=-\gamma z - \ln z +\sum_{k=1}^{\infty}\left(-\ln\left(1+\frac{z}{k}\right)+\frac{z}{k}\right$$ and thus, taking derivatives,$$\psi(z)=-\gamma-\frac{1}{z}+\sum_{k=1}^{\infty}\left(-\frac{1/k}{1+\frac{z}{k}}+\frac{1}{k}\right) =-\gamma-\frac{1}{z}+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{z+k}\right$$ The second formula follows after rearranging terms (the rearrangement is legal since we are simply exchanging adjacent terms, so partial sums remain the same).