Hurwitz zeta function

Remarks

For $a = 1$, the series is $\sum_{n \ge 1} n^{-s}$, so $\zeta(s, 1)$ is the Riemann zeta function $\zeta(s)$. For $a = 1/2$, the terms are $2^s (2n + 1)^{-s}$, so $\zeta(s, 1/2) = 2^s \sum_{m \text{ odd}} m^{-s} = (2^s - 1)\zeta(s)$. The value $s = 1$ is not allowed in the series: there the terms behave like $1/n$ and the series diverges. The continuation has a single pole, at $s = 1$; it is simple with residue $1$. For a Dirichlet character $\chi$ modulo $q$, grouping $n$ by its residue gives $$L(s, \chi) = q^{-s} \sum_{r=1}^{q} \chi(r)\, \zeta\!\left(s, \frac{r}{q}\right).$$ Some authors also allow complex $a$ with $\operatorname{Re} a > 0$.