Polygamma function

Remarks

The series $\psi(z) = -\gamma + \sum_{k=0}^\infty \left(\frac{1}{k+1} - \frac{1}{z+k}\right)$, with $\gamma$ the Euler–Mascheroni constant, can be differentiated term by term. For $m \ge 1$ this gives $$\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}} = (-1)^{m+1}\, m!\, \zeta(m+1, z),$$ where $\zeta(s, z)$ is the Hurwitz zeta function.

Example: for $m = 2$ and $z = 1$ the series gives $\psi^{(2)}(1) = -2 \sum_{k \ge 1} k^{-3} = -2\zeta(3)$.

Non-example of a valid argument: at $z = -n$ with $n \ge 0$ an integer the term $k = n$ of the series is infinite. There $\psi^{(m)}$ has a pole of order $m + 1$.

The function $\psi^{(0)} = \psi$ is the digamma function and $\psi^{(1)}$ is the trigamma function. The recurrence $\psi^{(m)}(z+1) = \psi^{(m)}(z) + (-1)^m\, m!\, z^{-m-1}$ follows by differentiating $\psi(z+1) = \psi(z) + 1/z$ $m$ times.