proof of Neumann series in Banach algebras

Let x be an element of a Banach algebra with identity, . By applying the properties of the Norm in a Banach algebra, we see that the partial sums form a Cauchy sequence: $\parallel {\sum }_{n=l}^m{x}^n\parallel \le {\sum }_{n=l}^m{\parallel x\parallel }^n\to 0$ for $l,m\to \mathrm{\infty }$ (as is well known from real analysis), so by completeness of the Banach Algebra, the series converges to some element $y={\sum }_{n=0}^{\mathrm{\infty }}{x}^n$.

We observe that for any $m\in \mathbb{N}$,

$$(1-x)\sum _{n=0}^m{x}^n=\sum _{n=0}^m{x}^n-\sum _{n=1}^{m+1}{x}^n=1-x^{m+1}$$(1)

Furthermore, $\parallel x^{m+1}\parallel \le {\parallel x\parallel }^{m+1}$, so ${lim}_m{x}^{m+1}=0$.

Thus, by taking the limit $m\to \mathrm{\infty }$ on both sides of (1), we get

$$(1-x)y=1$$

(We can exchange the limit with the multiplication by $(1-x)$, since the multiplication in Banach algebras is continuous)

Since the Banach algebra generated by a single element is commutative and $(1-x)$ and $y$ are both in the Banach algebra generated by $x$, we also get $y(1-x)=1$. Hence, $y={(1-x)}^{-1}$.

As in the first paragraph, the last claim $y\le \frac{1}{1-\parallel y\parallel }$ again follows by applying the geometric series for real numbers.