= Solution
Insert $u=-1/x^2$ into the <exponential series> $e^u=\sum_{n=0}^\infty u^n/n!$. For every fixed $x\ne0$, this gives the expansion with <absolute convergence>
$$
\boxed{f(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\,x^{2n}},\qquad a_n=\frac{(-1)^n}{n!}.}
$$
This is a <Laurent series> in inverse powers of $x$, as requested. It is not a <Maclaurin series> about zero: that series has every coefficient zero and cannot equal $f$ on a punctured neighborhood of zero. There is no contradiction, since the inverse-power expansion does not apply at $x=0$.
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