Where is the mistake with my proof that $\sum\limits_{m=0}^{a-1}\operatorname{Li}_s(ze^{\frac{2\pi im }{a}})=a^{1-s}\operatorname{Li}_s(z^a)$

Question

I tried to prove that $$\sum_{m=0}^{a-1}\operatorname{Li}_s(ze^{\frac{2\pi im }{a}})=a^{1-s}\operatorname{Li}_s(z^a)$$

where $\operatorname{Li}_s(x)$ is Polylogarithm function.

$$\sum_{m=0}^{a-1}\operatorname{Li}_s(ze^{\frac{2\pi im }{a}})=\sum_{n=1}^{\infty}\frac{z^n}{n^s}\sum_{m=0}^{a-1}e^{\frac{2\pi imn }{a}}$$

$$\sum_{m=0}^{a-1}\operatorname{Li}_s(ze^{\frac{2\pi im }{a}})=\sum_{n=1}^{\infty}\frac{z^n}{n^s}\frac{1-e^{2\pi in}}{1-e^{\frac{2\pi in }{a}}}=0$$

so where is the mistake and any way how to prove this formula ?

Answer 1

so where is the mistake

If $n$ is a multiple of $a$, then

$$\sum_{m=0}^{a-1} \exp \left(\frac{2\pi i mn}{a}\right) = \sum_{m=0}^{a-1} 1 = a.$$