partial fraction series for digamma function

Theorem 1

$$\psi (z)=-\gamma -\frac{1}{z}+\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{k}-\frac{1}{z+k}\right)=-\gamma +\sum _{k=0}^{\mathrm{\infty }}\left(\frac{1}{k+1}-\frac{1}{z+k}\right)$$

Proof:Start with

$$\mathrm{\Gamma }(z)=\frac{e^{-\gamma z}}{z}\prod _{k=1}^{\mathrm{\infty }}{\left(1+\frac{z}{k}\right)}^{-1}{e}^{z/k},$$

so

$$\ln \mathrm{\Gamma }(z)=-\gamma z-\ln z+\sum _{k=1}^{\mathrm{\infty }}\left(-\ln \left(1+\frac{z}{k}\right)+\frac{z}{k}\right)$$

and thus, taking derivatives,

$$\psi (z)=-\gamma -\frac{1}{z}+\sum _{k=1}^{\mathrm{\infty }}\left(-\frac{1/k}{1+\frac{z}{k}}+\frac{1}{k}\right)=-\gamma -\frac{1}{z}+\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{k}-\frac{1}{z+k}\right)$$

The second formula follows after rearranging terms (the rearrangement is legal since we are simply exchanging adjacent terms, so partial sums remain the same).