= Solution
The <geometric series> for $(2+z)^{-1}$ converges for $|z|<2$, while the cancelled <sine> terms have an everywhere convergent <Taylor series>. Hence the requested <Laurent series> has no negative powers:
$$
\boxed{f(z)=\sum_{n=0}^\infty\frac{(-1)^n z^n}{2^{n+1}}+\sum_{k=1}^\infty\frac{(-1)^k z^{2k-1}}{(2k+1)!},\qquad 0<|z|<2.}
$$
For example,
$$
f(z)=\frac12-\frac{5z}{12}+\frac{z^2}{8}-\frac{13z^3}{240}+\frac{z^4}{32}+O(z^5).
$$
The extension through zero also makes this a <Taylor series>. The <pole> at $-2$ fixes its <radius of convergence> at two.
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