[proofplan]
The strategy is the Banach-algebra analogue of the geometric series. Setting $x = 1 - a$, the hypothesis $\|x\| < 1$ together with submultiplicativity of the norm forces the partial sums of $\sum_{n=0}^\infty x^n$ to be Cauchy in $A$. Completeness of $A$ gives a sum $s \in A$. Telescoping the product $a \cdot s = (1 - x) \sum_{n=0}^N x^n = 1 - x^{N+1}$ and passing to the limit identifies $s$ as a two-sided inverse of $a$. The norm bound is the geometric-series estimate applied to $\|s\|$.
[/proofplan]
[step:Reduce to a Neumann series for $x := 1 - a$]
Set $x := 1 - a \in A$, so that $a = 1 - x$ and $\|x\| < 1$ by hypothesis. Define the partial sums
\begin{align*}
s_N : \mathbb{N} &\to A \\
N &\mapsto \sum_{n=0}^N x^n,
\end{align*}
where $x^0 := 1$. We will show $(s_N)_{N \geq 0}$ converges in $A$ and identify the limit as $a^{-1}$.
[/step]
[step:Show absolute convergence of $\sum_{n \geq 0} x^n$ via submultiplicativity]
We claim the series $\sum_{n=0}^\infty x^n$ converges absolutely in $A$, i.e.\ $\sum_{n=0}^\infty \|x^n\| < \infty$. The Banach-algebra norm satisfies $\|uv\| \leq \|u\|\|v\|$ for all $u, v \in A$ (submultiplicativity), so by induction on $n$,
\begin{align*}
\|x^n\| \leq \|x\|^n \quad \text{for all } n \geq 0.
\end{align*}
Since $r := \|x\| < 1$ by hypothesis, the geometric series $\sum_{n=0}^\infty r^n = \frac{1}{1 - r}$ converges, and the comparison test gives
\begin{align*}
\sum_{n=0}^\infty \|x^n\| \leq \sum_{n=0}^\infty r^n = \frac{1}{1 - r} < \infty.
\end{align*}
Hence $\sum_{n=0}^\infty x^n$ is absolutely convergent in $A$.
[guided]
We want to make sense of $\sum_{n=0}^\infty x^n$ as an element of $A$. The standard tool is: in a Banach space, an absolutely convergent series converges. So we need $\sum \|x^n\| < \infty$.
The norm on a Banach algebra is required to satisfy submultiplicativity, $\|uv\| \leq \|u\|\|v\|$. This is the axiom that makes Banach algebras different from Banach spaces — it links the multiplicative and the metric structures. Iterating,
\begin{align*}
\|x^n\| = \|x \cdot x^{n-1}\| \leq \|x\| \cdot \|x^{n-1}\| \leq \dots \leq \|x\|^n,
\end{align*}
where the induction starts at $n = 0$ with $\|x^0\| = \|1\| \leq 1$ in the standard convention (or with $\|x^0\| = \|1\|$ in any case — the bound $\|x^n\| \leq \|x\|^n$ holds for $n \geq 1$ by induction, and we may bound the $n = 0$ term separately by $\|1\|$).
With $r := \|x\| < 1$, the real geometric series $\sum_{n=0}^\infty r^n = \frac{1}{1 - r}$ converges. Therefore
\begin{align*}
\sum_{n=0}^\infty \|x^n\| \leq \sum_{n=0}^\infty r^n = \frac{1}{1 - r} < \infty,
\end{align*}
so the series is absolutely convergent.
[/guided]
[/step]
[step:Pass from absolute convergence to convergence in the Banach algebra $A$]
Absolute convergence of $\sum x^n$ implies convergence in $A$ because $A$ is complete. Concretely, for $M > N$,
\begin{align*}
\|s_M - s_N\| = \Big\| \sum_{n=N+1}^M x^n \Big\| \leq \sum_{n=N+1}^M \|x^n\| \leq \sum_{n=N+1}^M \|x\|^n,
\end{align*}
and the right-hand side is a tail of a convergent geometric series, hence tends to $0$ as $N \to \infty$. So $(s_N)$ is Cauchy in $A$. Since $A$ is a Banach space (in particular, complete), $(s_N)$ converges to some $s \in A$:
\begin{align*}
s := \lim_{N \to \infty} s_N = \sum_{n=0}^\infty x^n.
\end{align*}
[/step]
[step:Telescope $(1 - x) s_N$ to identify $s$ as a two-sided inverse of $a$]
We compute
\begin{align*}
a \cdot s_N = (1 - x) \sum_{n=0}^N x^n = \sum_{n=0}^N x^n - \sum_{n=0}^N x^{n+1} = \sum_{n=0}^N x^n - \sum_{n=1}^{N+1} x^n = x^0 - x^{N+1} = 1 - x^{N+1},
\end{align*}
using distributivity of multiplication over addition in $A$ (a ring axiom). Similarly,
\begin{align*}
s_N \cdot a = \sum_{n=0}^N x^n (1 - x) = 1 - x^{N+1}.
\end{align*}
Now $\|x^{N+1}\| \leq \|x\|^{N+1} \to 0$ as $N \to \infty$ since $\|x\| < 1$, so $x^{N+1} \to 0$ in $A$. Multiplication in a Banach algebra is jointly continuous: if $u_N \to u$ and $v_N \to v$, then $u_N v_N \to uv$ (this follows from $\|u_N v_N - uv\| \leq \|u_N\| \|v_N - v\| + \|u_N - u\| \|v\|$ and boundedness of $(u_N)$). Applying this with $u_N = a$ (constant) and $v_N = s_N \to s$:
\begin{align*}
a s = \lim_{N \to \infty} a s_N = \lim_{N \to \infty} (1 - x^{N+1}) = 1.
\end{align*}
Symmetrically, $s a = 1$. Therefore $a$ is invertible in $A$ with $a^{-1} = s$.
[/step]
[step:Establish the norm bound via the geometric-series estimate]
The norm is continuous on $A$ (as it is Lipschitz with constant $1$: $|\|u\| - \|v\|| \leq \|u - v\|$). Hence
\begin{align*}
\|a^{-1}\| = \|s\| = \Big\| \lim_{N \to \infty} s_N \Big\| = \lim_{N \to \infty} \|s_N\|.
\end{align*}
By the triangle inequality and the bound $\|x^n\| \leq \|x\|^n$ from Step 2,
\begin{align*}
\|s_N\| = \Big\| \sum_{n=0}^N x^n \Big\| \leq \sum_{n=0}^N \|x^n\| \leq \sum_{n=0}^N \|x\|^n.
\end{align*}
Letting $N \to \infty$ in the right-hand side, which converges since $\|x\| < 1$,
\begin{align*}
\|a^{-1}\| \leq \sum_{n=0}^\infty \|x\|^n = \frac{1}{1 - \|x\|} = \frac{1}{1 - \|1 - a\|}.
\end{align*}
This is the asserted bound, completing the proof.
[/step]
