Let $\mathcal{A}$ be a unital commutative Banach algebra over the complex numbers. Fix an element $a \in \mathcal{A}$. I am working with the resolvent set $\{\lambda \in \mathbb{C}: \lambda 1_{\mathcal{A}} - a$ is invertible$\}$. Define the resolvent function (omitting the indicator from now on):
$$ R(a,\lambda) = (\lambda - a)^{-1} = \lambda^{-1} (1 - \frac{a}{\lambda})^{-1} = \lambda^{-1}\sum _{n=0}^\infty (\frac{a}{\lambda})^n$$
I am trying to complete my lecture notes which go as follows:
We know that the above converges for $||a||<|\lambda| $ so that $||\frac{a}{\lambda}||<1$. Observe that we proved for all $\lambda \in \sigma (a)$, where $\sigma(a)$ is the spectrum of $a$, $|\lambda| \leq ||a||$. Therefore, $r(a) \leq ||a||$, where $r(a)$ is spectral radius. We want to show that the above converges for $|\lambda| \geq r(a)$.
To do that, we again evaluate resolvent function at infinity, $\lambda = \frac{1}{z}$.
$$ R(a,\frac{1}{z}) = z \sum a^n z^n $$
We know that the above converges for (equivalently) $|z| < 1/||a||$. We want to show it converges for $|1/z| = |\lambda|\geq r(a)$, that is $|z| < 1/r(a) \geq 1/||a||$.
if this somehow helps, we also showed that $$ \lambda1_\mathcal{A}-a=(z_0-a)-(z_0-\lambda)=(z_0-a)(1_\mathcal{A}-\frac{z_0-\lambda}{z_0-a})$$ We know the the latter term is invertible if $\|\frac{z_0-\lambda}{z_0-a}\|<1$ $$ (\lambda-a)^{-1}=\sum_{n=0}^\infty\left(\frac{z_0-\lambda}{z_0-a}^n \right)(z_0-a)^{-1} $$
Then my Professor said to just accept the fact that $R(a,\frac{1}{z})$ is holomorphic, and recall the following result from complex analysis: For a holomorphic function defined on some open set $O$, and for any $z_0 \in O$, the power series of $f$ about $z_0$ will converge in the largest open disk about $z_0$.
In the next lecture we also worked with the map $\lambda \rightarrow \lambda^{-1}\phi(a^n)\lambda^{-n}$, where $\phi \in \hat{\mathcal{A}}$, but that was to show that $r(a) = \lim_{n} ||a^n||^{1/n}$. But maybe it is relevant, I was very confused about that proof also.
Please help me finish the proof that the resolvent converges for $|\lambda| \geq r(a)$.