Let $X$ be a $\mathbb C$-Banach space, $A\in\mathfrak L(X)$ and $\rho(A)$ denote the resolvent set of $A$. It is easy to prove that $\sum_{n=0}^\infty\frac{A^n}{\lambda^n}$ is absolutely convergent with $$\lambda R_\lambda(A)=\sum_{n\in\mathbb N_0}\frac{A^n}{\lambda^n}\tag1$$ for all $\lambda\in\mathbb C\setminus\overline B_{r(A)}(0)$, where $r(A)$ denotes the spectral radius of $A$.
I would like to ask this question in a more general setting, but I wasn't sure how I would need to formulate this setting. However, what I would like to know is whether we can immediately conclude that $$f(\lambda):= R_\lambda(A)=(\lambda-A)^{-1}$$ is analytic; just from $(1)$ and knowing that the series converges absolutely.
I've read that this is true, but I don't know why. It's not that I'm not able to obtain the result: If $\lambda_0\in\rho(A)$, we can write $$\lambda-A=(\lambda_0-A)(1-\underbrace{(\lambda_0-\lambda)R_{\lambda_0}(A)}_{=:\:B_\lambda}\;\;\;\text{for all }\lambda\in\mathbb C\tag2.$$ Now if $\lambda\in\mathbb C$ with $\left\|B_\lambda\right\|_{\mathfrak L(X)}<1$, i.e. $$|\lambda_0-\lambda|<\varepsilon:=\left\|R_{\lambda_0}(A)\right\|_{\mathfrak L(X)}^{-1}\tag3,$$ we know from a basic result on Neumann series that $1-B_\lambda$ is bijective and $$(1-B_\lambda)^{-1}=\sum_{n\in\mathbb N_0}B_\lambda^n\tag4.$$ In particular, $\lambda-A$ is bijective (hence $\lambda\in\rho(A)$) and $$R_\lambda(A)=\sum_{n\in\mathbb N_0}(\lambda_0-\lambda)^nR_{\lambda_0}(A)^{n+1}\tag5.$$
So, I really just want to know whether the desired claim holds more generally; without the need of the extra arguments I've provided above.
Let $T:=\lambda_0-A$ and $z:=\lambda-\lambda_0$; then $\lambda-A=T+z$.
Then for small $z$, by the usual binomial expansion, $$(T+z)^{-1}=T^{-1}-zT^{-2}+z^2T^{-3}+\cdots$$ The higher order terms are of order $o(z)$ since $$\|\sum_{n=2}^\infty (-z)^nT^{-n-1}\|\le\sum_{n=2}^\infty|z|^n\|T^{-1}\|^{n+1}=\frac{|z|^2\|T^{-1}\|^3}{1-|z|\|T^{-1}\|}=o(z)$$ Thus $T=\lambda_0-A$ is complex-differentiable, i.e., analytic, in $\lambda_0$.