[Sage] residue

hbghlyj

$$z-\sin z={z^3\over6}\left(1-{z^2\over20}+O(z^4)\right)$$ Therefore $${1\over z-\sin z}={6\over z^3}\left(1+{z^2\over20}+O(z^4)\right)=6{1\over z^3}+{3\over10}{1\over z}+O(z)$$ so the residue at 0 is ${3\over10}$.