[proofplan]
We first use submultiplicativity of the Banach algebra norm to dominate $\|a^n\|_A$ by the scalar geometric sequence $\|a\|_A^n$. This makes the sequence of partial sums of $\sum_{n=0}^{\infty}a^n$ Cauchy, hence convergent by completeness of $A$. The finite partial sums satisfy an exact telescoping identity with $1_A-a$, and the remainder $a^{N+1}$ tends to $0$. Passing this identity to the norm limit gives a two-sided inverse for $1_A-a$.
[/proofplan]
[step:Bound the powers of $a$ by a scalar geometric sequence]
Let $r:=\|a\|_A$. By hypothesis, $0\le r<1$. Define $a^0:=1_A$, and for each $n\in\mathbb N$ define $a^n$ recursively by $a^n:=a^{n-1}a$.
We claim that
\begin{align*}
\|a^n\|_A\le r^n
\end{align*}
for every integer $n\ge 0$. For $n=0$, this reads $\|1_A\|_A\le 1$ if the Banach algebra is normalized. If the chosen convention only assumes $\|1_A\|_A\ge 1$, the convergence argument below uses the tail estimates for $n\ge 1$, and the single term $a^0=1_A$ is irrelevant to convergence. For $n\ge 1$, submultiplicativity gives inductively
\begin{align*}
\|a^n\|_A=\|a^{n-1}a\|_A\le \|a^{n-1}\|_A\|a\|_A\le r^{n-1}r=r^n.
\end{align*}
[/step]
[step:Construct the norm limit of the Neumann series]
For each integer $N\ge 0$, define the partial sum
\begin{align*}
s_N:=\sum_{n=0}^{N}a^n\in A.
\end{align*}
We show that $(s_N)_{N\ge 0}$ is Cauchy in the norm $\|\cdot\|_A$. If $M>N\ge 0$, then the triangle inequality and the bound from the previous step give
\begin{align*}
\|s_M-s_N\|_A\le \sum_{n=N+1}^{M}\|a^n\|_A\le \sum_{n=N+1}^{M}r^n.
\end{align*}
Since $0\le r<1$, the scalar geometric series $\sum_{n=0}^{\infty}r^n$ converges in $\mathbb R$, so its tails tend to $0$. Hence $(s_N)_{N\ge 0}$ is Cauchy in $A$. Because $A$ is complete, there exists an element $s\in A$ such that
\begin{align*}
\lim_{N\to\infty}\|s_N-s\|_A=0.
\end{align*}
By definition, this element is the norm-convergent series
\begin{align*}
s=\sum_{n=0}^{\infty}a^n.
\end{align*}
[guided]
The goal of this step is to make the infinite expression $\sum_{n=0}^{\infty}a^n$ legitimate as an element of $A$. We do this by proving that its partial sums converge in the [Banach space](/page/Banach%20Space) norm.
For each integer $N\ge 0$, define
\begin{align*}
s_N:=\sum_{n=0}^{N}a^n\in A.
\end{align*}
To prove convergence in a Banach space, it is enough to prove that the sequence of partial sums is Cauchy, because completeness of $A$ then supplies a norm limit. Let $M>N\ge 0$. Subtracting the two finite sums gives
\begin{align*}
s_M-s_N=\sum_{n=N+1}^{M}a^n.
\end{align*}
Applying the triangle inequality in the [normed vector space](/page/Normed%20Vector%20Space) $A$, and then using the power estimate $\|a^n\|_A\le r^n$ with $r=\|a\|_A$, gives
\begin{align*}
\|s_M-s_N\|_A\le \sum_{n=N+1}^{M}\|a^n\|_A\le \sum_{n=N+1}^{M}r^n.
\end{align*}
The right-hand side is a tail of the scalar geometric series with ratio $r$, and the hypothesis $\|a\|_A<1$ is exactly the condition $0\le r<1$. Therefore the scalar tails tend to $0$ as $N\to\infty$, uniformly in $M>N$. Hence $(s_N)_{N\ge 0}$ is Cauchy in $A$.
Since $A$ is a Banach algebra, its underlying normed [vector space](/page/Vector%20Space) is complete. Thus there exists $s\in A$ such that
\begin{align*}
\lim_{N\to\infty}\|s_N-s\|_A=0.
\end{align*}
This element $s$ is, by definition, the norm sum of the Neumann series:
\begin{align*}
s=\sum_{n=0}^{\infty}a^n.
\end{align*}
[/guided]
[/step]
[step:Verify the finite telescoping inverse identities]
For each integer $N\ge 0$, associativity and distributivity in the algebra give
\begin{align*}
(1_A-a)s_N=(1_A-a)\sum_{n=0}^{N}a^n.
\end{align*}
Expanding the finite product yields
\begin{align*}
(1_A-a)s_N=\sum_{n=0}^{N}a^n-\sum_{n=0}^{N}a^{n+1}=1_A-a^{N+1}.
\end{align*}
The same computation on the other side gives
\begin{align*}
s_N(1_A-a)=\sum_{n=0}^{N}a^n-\sum_{n=0}^{N}a^{n+1}=1_A-a^{N+1}.
\end{align*}
Thus
\begin{align*}
(1_A-a)s_N=s_N(1_A-a)=1_A-a^{N+1}
\end{align*}
for every integer $N\ge 0$.
[/step]
[step:Pass the finite identities to the norm limit]
The estimate from the first step gives
\begin{align*}
\|a^{N+1}\|_A\le r^{N+1}.
\end{align*}
Since $0\le r<1$, we have $r^{N+1}\to 0$, hence
\begin{align*}
\lim_{N\to\infty}\|a^{N+1}\|_A=0.
\end{align*}
Multiplication by a fixed element is norm-continuous in a normed algebra. Indeed, for any sequence $(x_N)_{N\ge 0}$ in $A$ with $x_N\to x$ in $\|\cdot\|_A$ and any fixed $b\in A$,
\begin{align*}
\|bx_N-bx\|_A=\|b(x_N-x)\|_A\le \|b\|_A\|x_N-x\|_A\to 0,
\end{align*}
and similarly
\begin{align*}
\|x_Nb-xb\|_A=\|(x_N-x)b\|_A\le \|x_N-x\|_A\|b\|_A\to 0.
\end{align*}
Applying this with $b=1_A-a$ and $x_N=s_N$, $x=s$, we obtain
\begin{align*}
(1_A-a)s=\lim_{N\to\infty}(1_A-a)s_N.
\end{align*}
Using the finite identity and $a^{N+1}\to 0$, this gives
\begin{align*}
(1_A-a)s=\lim_{N\to\infty}(1_A-a^{N+1})=1_A.
\end{align*}
The same continuity argument on the right gives
\begin{align*}
s(1_A-a)=\lim_{N\to\infty}s_N(1_A-a)=\lim_{N\to\infty}(1_A-a^{N+1})=1_A.
\end{align*}
Therefore $s$ is a two-sided inverse of $1_A-a$.
[/step]
[step:Identify the inverse with the Neumann series]
From the previous step,
\begin{align*}
(1_A-a)s=s(1_A-a)=1_A.
\end{align*}
Thus $1_A-a$ is invertible in $A$, and its inverse is the element $s$. Since $s$ was constructed as the norm limit of the partial sums of $\sum_{n=0}^{\infty}a^n$, we conclude that
\begin{align*}
(1_A-a)^{-1}=s=\sum_{n=0}^{\infty}a^n.
\end{align*}
This proves both the invertibility of $1_A-a$ and the asserted norm-convergent formula for its inverse.
[/step]
