[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*}[/step]

[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: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]