Laurent series for $\cosh(z)$

Question

In particular, since $\cosh(z) = \frac{e^{z}+e^{-z}}{2}$, and $e^{z} = \sum_{n=0}^{\infty} \frac{z^{n}}{n!}$, why is it not the case that $\cosh(z) = \frac{\sum_{n=0}^{\infty} \frac{z^{n}}{n!} + (\sum_{n=0}^{\infty} \frac{z^{n}}{n!})^{-1}}{2} = \frac{\sum_{n=0}^{\infty} \frac{z^{n}}{n!} + (\sum_{n=0}^{-\infty} \frac{z^{n}}{n!})}{2} = \frac{\sum_{-\infty}^{\infty} \frac{z^{n}}{n!}+1}{2}$?

Answer 1

$$e^{-z} = \sum_{n=0}^{\infty} \frac{(-z)^n}{n!} \neq \sum_{n=-\infty}^{0} \frac{z^n}{n!}.$$ Also, $n!$ is not defined for $n <0$.