$\lambda \in \rho(A)$ and $|\lambda - \mu| < \Vert R_\lambda \Vert^{-1}$ then $\sum_{n=0}^\infty (\lambda - \mu)^n R_\lambda ^{n+1} = (\mu - A)^{-1}$

Question

This is a remark from Ethier-Kurtz Markov Processes. I don't understand why the sum below is a bounded linear operator and in fact $(\mu - A)^{-1}$. Also, how does this imply that $\rho(A)$ is open? I would greatly appreciate any explanation. Above this remark, there is the resolvent identity $$R_\lambda R_\mu = R_\mu R_\lambda = (\lambda - \mu)^{-1}(R_\mu - R_\lambda).$$

Let $A$ be a closed linear operator on $L$. If $\lambda \in \rho(A)$, the resolvent set of $A$, and $|\lambda - \mu| < \Vert R_\lambda \Vert^{-1}$, where $R_\lambda = (\lambda I-A)^{-1}$, then $$\sum_{n=0}^\infty (\lambda - \mu)^n R_\lambda ^{n+1}$$ defines a bounded linear operator that is in fact $(\mu - A)^{-1}$. In particular, this implies that $\rho(A)$ is open in $\mathbb{R}$.

Answer 1

Τhe operator $T=\sum_0^\infty (\lambda-\mu)^n R_\lambda^{n+1}$ is a bounded linear operator. Indeed this a Newman series, of the form $\sum a_nR^n$, and convergence is guaranteed whenever $$ \|R\| \limsup_{n\to\infty}|a_n|^{1/n}<1. $$ It suffices to show that $(\mu-A)Tx=x$, for all $x$ is our Banach space.

We have that $$ (\mu-A)Tx=\big((\lambda-A)-(\lambda-\mu)\big)\sum_0^\infty (\lambda-\mu)^n R_\lambda^{n+1}x \\=\sum_0^\infty (\lambda-\mu)^{n} (\lambda-A)R_\lambda^{n+1}x-\sum_0^\infty (\lambda-\mu)^{n+1} R_\lambda^{n+1}x \\=\sum_0^\infty (\lambda-\mu)^{n} R_\lambda^{n}x-\sum_0^\infty (\lambda-\mu)^{n+1} R_\lambda^{n+1}x=(\lambda-\mu)^0R_\lambda^0x=x. $$ Note that $(\lambda-A)R^{n+1}_\lambda=(\lambda-A)(\lambda-A)^{-1}R^{n}_\lambda=R^{n}_\lambda.$