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I'm trying to find the Laurent expansion for

$$\frac{e^{1/z^2}}{z - 1}$$ about $z_0 = 0$.

Writing the series for $e^{1/z^2}$ and $1/(z-1)$ individually gives

$$\frac{e^{1/z^2}}{z - 1} = -\left(\sum_{n=0}^\infty z^n \right) \left(\sum_{n=0}^\infty \frac{1}{z^{2n}n!} \right)$$ How do I multiply these two series, and under what conditions is the convergence of the product guaranteed?

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  • $\begingroup$ your two series converge absolutely on $0 < |z| < 1$ so you can multiply them and change the order of summation, it will still be absolutely convergent. And be careful, there are two posible Laurent series at $z=0$ for your function, the one you wrote converging on the annulus $0 < |z| < 1$, and the other converging on the annulus $1 < |z| < \infty$ $\endgroup$ – reuns Jul 24 '16 at 16:53
  • $\begingroup$ and for the multiplication : $\displaystyle\left(\sum_{n=0}^\infty z^n \right) \left(\sum_{m=0}^\infty \frac{1}{z^{2m}m!} \right) = \sum_{k=-\infty}^\infty z^k \underbrace{\sum_{n \ge 0,m \ge 0, \ n - 2m = k} \frac{1}{m!}}_{\text{convolution of the coefs}} = \ldots$ $\endgroup$ – reuns Jul 24 '16 at 16:56

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