Polylogarithm

Remarks

The series has radius of convergence $1$ for every $s$, because $\abs{k^{-s}}^{1/k} = k^{-\operatorname{Re} s/k} \to 1$.

Example: for $s = 1$ the series is $-\ln(1 - z)$, and for $s = 0$ and $s = -1$ it gives $\operatorname{Li}_0(z) = \frac{z}{1 - z}$ and $\operatorname{Li}_{-1}(z) = \frac{z}{(1 - z)^2}$. For $\operatorname{Re} s > 1$ the series also converges at $z = 1$, and $\operatorname{Li}_s(1) = \zeta(s)$, the Riemann zeta function.

Non-example of a valid argument: for $s = 1$ the series diverges at $z = 1$ (it is the harmonic series), and $-\ln(1 - z)$ has a singularity there.

Termwise differentiation gives $z \frac{d}{dz} \operatorname{Li}_s(z) = \operatorname{Li}_{s-1}(z)$, so $\operatorname{Li}_{s+1}(z) = \int_0^z \frac{\operatorname{Li}_s(t)}{t}\, dt$. For $s = 2$ this is the dilogarithm.