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).

Title	partial fraction series for digamma function
Canonical name	PartialFractionSeriesForDigammaFunction
Date of creation	2013-03-22 16:23:40
Last modified on	2013-03-22 16:23:40
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 33B15
Classification	msc 30D30