Residue of a product of series

Question

I need to find the residue of $f(z)=\frac{e^{\frac{1}{z}}}{z-1}$ in $z=0$.

To do this, I proceeded to find the Laurent series of $f$ which is:

$\sum_{n=0}^{\infty} \frac{z^{-n}}{n!}\sum_{k=0}^{\infty} z^{k}$

Now I need the factor which multiplies $z^{-1}$ but I don't know how to operate with a product of series that have different expressions for the power of $z$

Answer 1

The coefficient of $z^{-1}$ in $\left(\sum_{n=0}^\infty \frac{z^{-n}}{n!}\right) \left(\sum_{k=0}^\infty z^k \right)$ comes from multiplying terms $z^{-n}/n!$ and $z^{n-1}$. Thus the coefficient is $$\sum_{n=1}^\infty \frac{1}{n!} = e - 1$$

Answer 2

Since $f$ is analytic in the neighborhood of $0$, its Laurent series there is actually a Taylor series and the residue is $0$.