Prove that $\mathrm{Li}_s(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^s}$ is a rational function.

Question

Polylogarithm: $$\mathrm{Li}_s(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^s}$$ Prove that $\mathrm{Li}_s(x)$ is a rational function if $s\in\mathbb{Z},\ s\leqslant 0$.

I tried to find some sort of regularity by evaluating polylogarithm for different $s$: $$ s=0:\ \ \mathrm{Li}_s(x)=\frac{x}{1-x}\\ s=-1:\ \ \mathrm{Li}_s(x)=\frac{x}{(1-x)^2}\\ s=-2:\ \ \mathrm{Li}_s(x)=\frac{x(x+1)}{(1-x)^3}\\ \vdots $$ However, from $s=-3$ there is no regularity: $$ s=-3:\ \ \mathrm{Li}_s(x)=\frac{x(x^2+4x+1)}{(1-x)^4}\\ \vdots $$

So, I would be glad if someone could tell me what I should actually do in this problem.

Answer 1

Hint: Show that

$$\frac{d}{dx}\text{Li}_n(x)=\frac{1}{x}\text{Li}_{n-1}(x)$$

then by induction if $\text{Li}_n(x)$ is rational, so is its derivative and hence $\text{Li}_{n-1}(x)$.

Answer 2

In case you're interested in a general formula, Wikipedia gives $$\begin{align} \mathrm{Li}_{-n}(x)&=\left(z\frac{\partial}{\partial z}\right)^n\frac{z}{1-z}\\ &=(-1)^{n+1}\sum_{k=0}^{n}\frac{\left\{{{n+1}\atop{k+1}}\right\}k!}{(1-z)^{k+1}}\\ &=\frac{1}{(1-z)^{n+1}}\sum_{k=0}^{n-1}\left<{{n}\atop{k}}\right>z^{n-k}, \end{align}$$ where $$\left\{{{n}\atop{k}}\right\}=\frac1{k!}\sum_{i=0}^{k}(-1)^i{k\choose i}(k-i)^n$$ are the Stirling numbers of the second kind, and $$\left<{{n}\atop{k}}\right>=\sum_{i=0}^{k}(-1)^i{{n+1}\choose i}(k+1-i)^n$$ are the Eularian numbers.