Let $z \neq 1$ be a complex number. Then
\begin{align}
\frac{1}{1-z} = \sum\_{n=0}^{\infty} z^n. 
\end{align}

We have 
\begin{align}
\frac{z^{-1}}{1-z^{-1}} = \sum\_{n=1}^{\infty} z^{-n}.
\end{align}

If we add the left hand sides of the two equations, then we have 
\begin{align}
\frac{1}{1-z} + \frac{z^{-1}}{1-z^{-1}} = 0.
\end{align}
If we add the right hand sides of the two equations, then we obtain 
\begin{align}
\sum\_{n = - \infty}^{\infty} z^n.
\end{align}
Therefore when $z \neq 1$, we have $\sum\_{z=-\infty}^{\infty} z^n= 0$. Do we have $\delta(z) = \sum\_{n=-\infty}^{\infty} z^n$? How to understand the delta function and analytic continuation? Any help will be greatly appreciated!