Let $x \in \mathcal{A}$ be an element of a unital Banach-algebra $\mathcal{A}$ and assume $\sigma(x) \subset \{ z \in \mathbf{C} : \vert 1 - z \vert < 1 \}$, where $\sigma(x)$ denotes the spectrum of the operator $y \mapsto x y$.
Now define
$$ \log (x) := - \sum_{n = 1}^\infty \frac{1}{n} (1 - x)^n$$
I'd like to show that the right hand side converges.
My approach: I thought on showing that $\sum_{n \geq 1} \frac{1}{n} \Vert (1 - x)^n \Vert < + \infty$, which suffices as $\mathcal{A}$ is a Banach-space.
One thing to note would be that
$$ \lim_{n \rightarrow \infty} \Vert (1 - x)^n \Vert^{\frac{1}{n}} = \sup_{\lambda \in \sigma (x)} \vert 1 - \lambda \vert < 1$$
since $\sigma(1 - x) = \{ 1 - \lambda : \lambda \in \sigma(x) \}$.
However, I think this doesn't suffice. So I thought on showing that $\Vert (1 - x)^n \Vert = \Vert 1 - x \Vert^n$ as one does for normal operators to get that the spectral radius is equal to the norm. But I doubt that this works in the general setting of a Banach-algebra.
Could anyone help me on how to prove this?
Thanks!