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Digamma function
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===Series formula=== Euler's product formula for the gamma function, combined with the functional equation and an identity for the Euler–Mascheroni constant, yields the following expression for the digamma function, valid in the complex plane outside the negative integers (Abramowitz and Stegun 6.3.16):<ref name="AbramowitzStegun"/> :<math>\begin{align} \psi(z + 1) &= -\gamma + \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n + z}\right), \qquad z \neq -1, -2, -3, \ldots, \\ &= -\gamma + \sum_{n=1}^\infty \left(\frac{z}{n(n + z)}\right), \qquad z \neq -1, -2, -3, \ldots. \end{align}</math> Equivalently, :<math>\begin{align} \psi(z) &= -\gamma + \sum_{n=0}^\infty \left(\frac{1}{n + 1} - \frac{1}{n + z}\right), \qquad z \neq 0, -1, -2, \ldots, \\ &= -\gamma + \sum_{n=0}^\infty \frac{z-1}{(n + 1)(n + z)}, \qquad z \neq 0, -1, -2, \ldots. \end{align}</math> ====Evaluation of sums of rational functions==== The above identity can be used to evaluate sums of the form : <math>\sum_{n=0}^\infty u_n=\sum_{n=0}^\infty \frac{p(n)}{q(n)},</math> where {{math|''p''(''n'')}} and {{math|''q''(''n'')}} are polynomials of {{mvar|n}}. Performing [[partial fraction]] on {{mvar|u<sub>n</sub>}} in the complex field, in the case when all roots of {{math|''q''(''n'')}} are simple roots, : <math>u_n=\frac{p(n)}{q(n)}=\sum_{k=1}^m \frac{a_k}{n+b_k}.</math> For the series to converge, :<math>\lim_{n\to\infty} nu_n=0,</math> otherwise the series will be greater than the [[harmonic series (mathematics)|harmonic series]] and thus diverge. Hence :<math>\sum_{k=1}^m a_k=0,</math> and :<math>\begin{align} \sum_{n=0}^\infty u_n &= \sum_{n=0}^\infty\sum_{k=1}^m\frac{a_k}{n+b_k} \\ &=\sum_{n=0}^\infty\sum_{k=1}^m a_k\left(\frac{1}{n+b_k}-\frac{1}{n+1}\right) \\ &=\sum_{k=1}^m\left(a_k\sum_{n=0}^\infty\left(\frac{1}{n+b_k}-\frac{1}{n+1}\right)\right)\\ &=-\sum_{k=1}^m a_k\big(\psi(b_k)+\gamma\big) \\ &=-\sum_{k=1}^m a_k\psi(b_k). \end{align}</math> With the series expansion of higher rank [[polygamma function]] a generalized formula can be given as :<math>\sum_{n=0}^\infty u_n=\sum_{n=0}^\infty\sum_{k=1}^m \frac{a_k}{(n+b_k)^{r_k}}=\sum_{k=1}^m \frac{(-1)^{r_k}}{(r_k-1)!}a_k\psi^{(r_k-1)}(b_k),</math> provided the series on the left converges.
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