Writing $e^{\frac{-1}{\epsilon}}$ as a sum $\sum_{n>0}a_n.\epsilon^n$

Question

given the $e^{\frac{-1}{\epsilon}}$, for $\epsilon>0.$

I want to write as an infinit sum : $$\sum_{n>0}a_n.\epsilon^n$$

All I can get is : $$\sum_{n>0}\frac{\epsilon^{-n}}{n!}$$

Answer 1

$f(z) = \displaystyle e^{-1/z}$ has an essential singularity at $z=0$, so it is not the sum of any Maclaurin series.

$f(z) \to 0$ as $z \to 0^+$ more rapidly than any nonzero Maclaurin series.

However, $f(z)$ does have a Laurent series, valid in the whole plane (except $0$):

$$ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\;z^{-n} $$