Consider $$f(z)=\frac{1}{e^z-1}$$ I want to expand it over $z_0=0$ in a Laurent series.
We know that $$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ And I know that $$\frac{1}{x-1}=-1-x-x^2-x^3-x^4-x^5+\cdots$$ But I can't get the correct answer for my desired series which is, $$\frac{1}{x}-\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}+O(x^6)$$ Any help?