False proof of 0=1 using Laurent series

Question

I found the following proof that 0 = 1: \begin{align*} \sum_{n=-\infty}^{\infty} 0\cdot z^n = 0 = \frac{1}{z-1} + \frac{1}{1-z} = \frac{1}{z}\frac{1}{1-\frac{1}{z}} + \frac{1}{1-z} \\ = \frac{1}{z} \sum_{n=0}^{\infty}\frac{1}{z^n} + \sum_{n=0}^{\infty}z^n = \sum_{n=-\infty}^{\infty} 1\cdot z^n \end{align*} from which it follows that 0 = 1. In every step, I can't find an error. If we take $f(z) = \frac{1}{z-1} + \frac{1}{1-z}$, then it is holomorfic on the disk around 0 with radius 1 (B(0,1)).The principal part $h(z)$ would be $\frac{1}{z-1}$, which is also holomorfic on $\mathbb{C}\backslash B[0,1]$ and also $h(z)$ tends to 0 if $|z|$ tends to $\infty$. So I can't see what is going wrong.

Answer 1

The equality $\dfrac{1}{z}\dfrac{1}{1-\frac{1}{z}} + \dfrac{1}{1-z} = \dfrac{1}{z} \sum \limits_{n=0}^{\infty}\dfrac{1}{z^n} + \sum \limits_{n=0}^{\infty}z^n$ only holds if $z\not \in \{0,1\}$, $\left|\frac 1 z\right|<1$ and $|z|<1$, i.e., never.