PartialFractionSeriesForDigammaFunction 2006-11-11 17:48:50 2007-03-13 17:13:38 Theorem digamma and polygamma function % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm}{Theorem} \begin{thm} \[\psi (z) = - \gamma -\frac{1}{z} + \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{z + k} \right)=-\gamma+\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{z+k}\right)\] \end{thm} \textbf{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).