TITLE: $e^{1/z}$ and Laurent expansion
QUESTION [15 upvotes]: $e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as
$$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$
The coefficients of this Laurent expansion are computed the same way as Taylor's. The question is how is that possible? If function is not holomoprhic at $z=0$, then it's not true that it is holomophic at $|z|<R$ and Taylor's coefficients can not be used. 
Please someone explain.

REPLY [22 votes]: I'm pretty sure most of former answers didn't get the key point.
The Laurent series expansion is defined on a "deleted neighborhood" around a singularity, in this case, $\{z: 0<\lvert z-0\rvert< R \}$. In this deleted neighborhood, $e^{1/z}$ is analytic. So for any point in this neighbourhood, we can expand $e^{z}$ first and then substitute $1/z$ in. 
As you can see in the Laurent expansion you gave, you can plug in any $z$ arbitrarily close to zero, calculate the infinite sum, and get a finite and well defined value. The laurent series doesn't give any information about the behavior of a function "on" its singularities.