Method for determining where Laurent series converge

Question

I have to find the Laurent series for $f(z) = \frac{1}{z^2(4z-1)}$. I know there are two series, centered at 0 and at 1/4, because that is where $f$ is not analytic, and I found the series using the Cauchy integral formula and the Residue theorem (They are $f(z)=-\frac{1}{z^2}-4\frac{1}{z}-16-64z+\dots$ and $f(z)=\frac{4}{z-1/4}-32+192(z-1/4)+\dots$ for 0 and 1/4, respectively), but how do I know where they converge? I know $f$ is analytic everywhere, except those two points, but that doesn't mean the series converges everywhere.