Theorem: Element of Unital Banach Algebra Close to Identity is Invertible

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {A, \norm \cdot}$ be a unital Banach algebra over $\Bbb F$ with identity element $\mathbf 1_A$. 
Let $a \in A$ be such that: 
:$\norm {\mathbf 1_A - a} < 1$ 
Then $a$ is invertible with inverse element $a^{-1}$ satisfying: 
:$\ds \norm {a^{-1} } \le \frac 1 {1 - \norm {\mathbf 1_A - a} }$

Proof:
Let:
:$x = \mathbf 1_A - a$
From Bound on Norm of Power of Element in Normed Algebra, we have: 
:$\norm {x^n} \le \norm x^n$
for each $n \in \Z_{\ge 0}$.
Then we have: 
:$\ds \sum_{n \mathop = 0}^\infty \norm {x^n} \le \sum_{n \mathop = 0}^\infty \norm x^n$ 
Since $\norm x < 1$, we have: 
:$\ds \sum_{n \mathop = 0}^\infty \norm x^n$ converges
from Sum of Infinite Geometric Sequence. 
So:
:$\ds \sum_{n \mathop = 0}^\infty \norm {x^n}$ converges.
Since $A$ is a Banach space, we have: 
:$\ds \sum_{n \mathop = 0}^\infty x^n$ converges.
by Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.
Let: 
:$\ds b = \sum_{n \mathop = 0}^\infty x^n$
We show that: 
:$\ds a b = \mathbf 1_A$
and:
:$\ds b a = \mathbf 1_A$
Note that: 
:$a = \mathbf 1_A - x$ 
Note that for each $N \in \N$, we have: 
{{begin-eqn}}
{{eqn	| l = \paren {\mathbf 1_A - x} \sum_{n \mathop = 0}^N x^n
		| r = \sum_{n \mathop = 0}^N \paren {\mathbf 1_A - x} x^n 
}}
{{eqn	| r = \sum_{n \mathop = 0}^N \paren {x^n - x^{n + 1} }
}}
{{eqn	| r = \mathbf 1_A - x^{N + 1}
}}
{{end-eqn}}
and similarly: 
{{begin-eqn}}
{{eqn	| l = \paren {\sum_{n \mathop = 0}^N x^n} \paren {\mathbf 1_A - x}
		| r = \sum_{n \mathop = 0}^N x^n \paren {\mathbf 1_A - x}
}}
{{eqn	| r = \sum_{n \mathop = 0}^N \paren {x^n - x^{n + 1} }
}}
{{eqn	| r = \mathbf 1_A - x^{N + 1}
}}
{{end-eqn}}
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {\mathbf 1_A - x^{N + 1} - \mathbf 1_A}
		| r = \norm {x^{N + 1} }
}}
{{eqn	| o = \le
		| r = \norm x^{N + 1}
}}
{{eqn	| o = \to
		| r = 0
		| c = since $\norm x < 1$
}}
{{end-eqn}}
so that: 
:$\mathbf 1_A - x^{N + 1} \to \mathbf 1_A$
from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence.
So taking $N \to \infty$ in: 
:$\ds a \sum_{n \mathop = 0}^N x^n = \mathbf 1_A - x^{N + 1}$
we obtain: 
:$a b = \mathbf 1_A$ 
from Product Rule for Sequence in Normed Algebra. 
Similarly taking $N \to \infty$ in: 
:$\ds \paren {\sum_{n \mathop = 0}^N x^n} a = \mathbf 1_A - x^{N + 1}$
we obtain: 
:$b a = \mathbf 1_A$
So $a$ is invertible with inverse element $b$. 
It remains to show that: 
:$\ds \norm b \le \frac 1 {1 - \norm x}$
For each $N \in \N$, we have that:
{{begin-eqn}}
{{eqn	| l = \norm {\sum_{n \mathop = 0}^N x^n}
		| o = \le
		| r = \sum_{n \mathop = 0}^N \norm {x^n}
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = \sum_{n \mathop = 0}^N \norm x^n
		| c = Bound on Norm of Power of Element in Normed Algebra
}}
{{eqn	| o = \le
		| r = \sum_{n \mathop = 0}^\infty \norm x^n
}}
{{eqn	| r = \frac 1 {1 - \norm x}
		| c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
Taking $N \to \infty$, we have: 
:$\ds \norm b \le \frac 1 {1 - \norm x} = \frac 1 {1 - \norm {\mathbf 1_A - a} }$ 
from Modulus of Limit: Normed Vector Space.
{{qed}}