Series for reciprocal of polylogarithm?

Question

The series representation of a polylogarithm of order $s$ is given by

$$\text{Li}_s(z) = \sum_{k=1}^{\infty}\frac{z^k}{k^s}$$

Are there any simplified expressions for $\dfrac{1}{\text{Li}_s(z)}$? Google doesn't seem to know.

Answer 1

I do not think that they are simplified forms in the reverse of a polylogarithm but you always can to build them by identification of series and series compositions. Just write $$\dfrac{1}{\text{Li}_s(z)}=\sum_{i=-1}^{\infty}a_i z^i$$ (the summation must start at $i=-1$ because the expansion of the polylogarithm starts with a first term equal to $z$) and so $$\sum_{i=-1}^{\infty}a_i z^i \sum_{k=1}^{\infty}\frac{z^k}{k^s}=1$$ and identify the $a_i$ as usual. I agree that the expressions are not very nice. I give you below a few of them $$\dfrac{1}{\text{Li}_1(z)}=\frac{1}{z}-\frac{1}{2}-\frac{z}{12}-\frac{z^2}{24}-\frac{19 z^3}{720}-\frac{3 z^4}{160}+O\left(z^5\right)$$ $$\dfrac{1}{\text{Li}_2(z)}=\frac{1}{z}-\frac{1}{4}-\frac{7 z}{144}-\frac{13 z^2}{576}-\frac{6911 z^3}{518400}-\frac{6151 z^4}{691200}+O\left(z^5\right)$$ $$\dfrac{1}{\text{Li}_3(z)}=\frac{1}{z}-\frac{1}{8}-\frac{37 z}{1728}-\frac{115 z^2}{13824}-\frac{1572859 z^3}{373248000}-\frac{2448719 z^4}{995328000}+O\left(z^5\right)$$ $$\dfrac{1}{\text{Li}_4(z)}=\frac{1}{z}-\frac{1}{16}-\frac{175 z}{20736}-\frac{865 z^2}{331776}-\frac{292581071 z^3}{268738560000}-\frac{771682411 z^4}{1433272320000}+O\left(z^5\right)$$

Trying to make this more general, we can find $$a_{-1}=1$$ $$a_0=-2^{-s}$$ $$a_1=2^{-2 s}-3^{-s}$$ $$a_2=-2^{-3 s}-2^{-2 s}+2^{1-s} 3^{-s}$$ $$a_3=2^{1-3 s}+2^{-4 s}-2^{-2 s} 3^{1-s}+3^{-2 s}-5^{-s}$$ $$a_4=-2^{-5 s}-3\ 2^{-4 s}-2^{-s} 3^{1-2 s}+2^{2-3 s} 3^{-s}+2^{1-2 s} 3^{-s}+2^{1-s} 5^{-s}-6^{-s}$$