Consider the following proof:
Theorem. Let $A$ be a unital Banach algebra and $a$ an element of $A$ such that $\Vert a \Vert < 1$. Then $1 - a \in \operatorname{Inv}(A)$ and $$(1-a)^{-1} = \sum_{n=0}^\infty a^n.$$ Proof. Since $$\sum_{n=0}^\infty \Vert a^n\Vert \leq \sum_{n=0}^\infty \Vert a \Vert^n = \left(1-\Vert a \Vert\right)^{-1} < + \infty,$$ the series $\sum_{n=0}^\infty a^n$ is convergent, to $b$ say, in $A$, and since $$\left(1-a\right) (1+\cdots+a^n) = 1 - a^{n+1}$$ converges to $\left(1-a\right) b = b \left(1-a\right)$ and to $1$ as $n \to \infty$, the element $b$ is the inverse of $1 - a$. $\square$
I am wondering about whether it is correct to argue as follows instead:
Since $\sum a^n$ is convergent the multiplication $a \sum_{n=0}^\infty a^n = \sum_{n=1}^\infty a^n$ and the subtraction $\sum_{n=0}^\infty a^n - \sum_{n=1}^\infty a^n = 1$ are justified. Therefore $\left(1-a\right)b = 1$.
Disclaimer: this is an answer to the above series of comments by Student. It was too long for a comment.
OK for "first" and "second". For "third", you shouldn't start by writing down $\frac1{1-a}$, since you precisely want to show that this exists. (By the way, the notation $\frac1{1-a}$ is usually "forbidden" in a vector-valued setting: it is generally agreed that one "must" write $(1-a)^{-1}$).
Likewise, you shouldn't write $\sum_0^\infty a_n$ as long as you don't know that the series is indeed convergent in $A$. So: start by proving that the series $\sum\Vert a^n\Vert$ is convergent (as you did). Then deduce that the series $\sum a^n$ is convergent in $A$ (because $A$ is a Banach algebra), and finally check as you did that $b=\sum_0^\infty a^n$ satisfies $(1-a)b=1=b(1-a)$, so that $1-a$ is indeed invertible with inverse $b$.
Final remark: part "second" is in fact never used.