when looking at Laurent Series expansions, I sometimes see something like: $$ \frac{1}{\sin z}= \frac{1}{z\left(1-\frac{z^2}{6}+\ldots\right)}=\frac{1}{z}\left(1+\frac{z^2}{6}+\frac{z^4}{120}+\ldots\right) $$
can anyone explain why $ \frac{1}{\left(1-\frac{z^2}{6}+\ldots\right)} $ is equal to $ (1+\frac{z^2}{6}+\frac{z^4}{120}+\ldots) $ ?
and what should I do in the general case of $ \frac{1}{\text{alternating-series}} $? I often see that it is evaluated to a positive series. why is that?
Your expansion is wrong: you can check with Mathematica that $$ \frac{1}{\sin z}=\frac{1}{z}\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+o(z^4)\right) $$ As I wrote in the comments you can use the two expansions: $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^5)$$ and $$(1+y)^{-1}=1-y+y^2+o(y^2)\ .$$ Now you have $$\frac{1}{\sin z}=\frac{1}{z}\left[1+\left(-\frac{z^2}{6}+\frac{z^4}{120}+o(z^4)\right)\right]^{-1}=$$ $$=\frac{1}{z}\left[1-\left(-\frac{z^2}{6}+\frac{z^4}{120}+o(z^4)\right)+\left(-\frac{z^2}{6}+\frac{z^4}{120}+o(z^4)\right)^2+o(z^4)\right]=$$ $$=\frac{1}{z}\left(1+\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^4}{36}+o(z^4)\right)\ ,$$ where in the last step I absorbed all term of order higher than 4 in $o(z^4)$. This is exactly the expansion I wrote at the beginning and the same you obtain with Mathematica or Wolfram Alpha.