When is it possible to multiply two Laurent series? For example, suppose I wanted to find the Laurent series for $\frac{e^{1/z}}{z-1}$ on the annulus $1<|z|<\infty$. On this annulus, we can write $$e^{1/z} = \sum_{n=0}^\infty \frac{1}{z^n \cdot n!}$$ and $$\frac{1}{z-1} = \frac{1}{z} \cdot \frac{1}{1-1/z} = \sum_{n=0}^\infty \frac{1}{z^{n+1}}$$ since $|1/z|<1$.
But I don't know if it is possible to just multiply the series term by term: $$ \frac{e^{1/z}}{z-1} = \sum_{n=0}^\infty \frac{1}{z^n \cdot n!} \cdot \sum_{n=0}^\infty \frac{1}{z^{n+1}}. $$
When the series involved are Taylor series converging on the same disk, they are both analytic functions and hence their product has a Taylor series on that same disk. It is easy to see that the coefficients of the Taylor series for the product (from Taylor's theorem) are the same as those found by formally multiplying the series term by term.
But in the case of my Laurent series above, both functions are analytic on the same annulus and hence the product has a Laurent series on that annulus. But the coefficients in Laurent's theorem are given by some contour integrals so it is not easy to see why they would be the same as those found by formal multiplication. I also remember reading somewhere that there may be convergence issues with multiplying Laurent series, but it didn't explain this in detail.