Lemma 33.2.2 (Neumann Series).label Let $A$ be a unital banach algebra and $x \in B_{A}(1, 1)$, then $x \in G(A)$ with
\[x^{-1}= \sum_{n = 0}^{\infty} (1 - x)^{n}\]
and $\normn{x^{-1}}_{A} \le (1 - \norm{x})_{A}^{-1}$.
Proof. Since $\norm{1 - x}_{A} < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^{\infty} (1 - x)^{n}$, then
\[(1 - x) \sum_{n = 0}^{\infty} (1 - x)^{n} = \sum_{n = 0}^{\infty} (1 - x)^{n} - 1 = \sum_{n = 0}^{\infty} (1 - x)^{n} (1 - x)\]
so $(1 - x)y = y - 1 = y(1 - x)$, and $xy = yx = 1$.$\square$
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