Proposition 6.1.16.label If $x\in A$ such that $\norm{x-1}{}<1$, then $x$ is invertible and
\begin{align*}x\inv=\sums{\infty}{n=0}(1-x)^{n}\end{align*}
where $a^{0}=1$ for $a\in A$ (here $a=1-x$)
Proof. Since $\abs{x-1}<1$, $\sums{\infty}{n=0}\norm{1-x}{}^{n}<\infty$, the series $\sums{\infty}{n=0}(1-x)^{n}$ converges in norm. Set $x'=\sums{\infty}{n=0}(1-x)^{n}$ then
\begin{align*}x'x&=xx'=(1-(1-x))x'=x'-(1-x)x'\\&=\sums{\infty}{n=0}(1-x)^{n}-\sums{\infty}{n=1}(1-x)^{n}\\&=1\end{align*}
$\square$