[proofplan]
Fix a point $\lambda_0\in\rho(T)$ and factor $T-\lambda I_X$ into the known invertible operator $T-\lambda_0 I_X$ times a small perturbation of the identity. The small factor is inverted by the Neumann series in the Banach algebra $\mathcal{L}(X)$ whenever $|\lambda-\lambda_0|\|R_T(\lambda_0)\|_{\mathcal{L}(X)}<1$. This gives both the inclusion of the corresponding disk in $\rho(T)$ and the displayed operator-norm [power series](/page/Power%20Series). Finally, the local power-series representation implies [complex differentiability](/page/Complex%20Differentiability), hence holomorphy, at every point of $\rho(T)$.
[/proofplan]
[step:Invert the small identity perturbation by a Neumann series]
Let $\lambda_0\in\rho(T)$ be fixed, and define
\begin{align*}
R_0:=R_T(\lambda_0)=(T-\lambda_0 I_X)^{-1}\in\mathcal{L}(X).
\end{align*}
Let $\lambda\in\mathbb C$ satisfy
\begin{align*}
|\lambda-\lambda_0|\|R_0\|_{\mathcal{L}(X)}<1.
\end{align*}
Define
\begin{align*}
A_\lambda:=(\lambda-\lambda_0)R_0\in\mathcal{L}(X).
\end{align*}
Then
\begin{align*}
\|A_\lambda\|_{\mathcal{L}(X)}\le |\lambda-\lambda_0|\|R_0\|_{\mathcal{L}(X)}<1.
\end{align*}
We prove the Neumann-series inversion in the Banach algebra $\mathcal{L}(X)$ directly. Set
\begin{align*}
S_\lambda:=\sum_{n=0}^{\infty}A_\lambda^n,
\end{align*}
where the series converges in $\mathcal{L}(X)$ because
\begin{align*}
\sum_{n=0}^{\infty}\|A_\lambda^n\|_{\mathcal{L}(X)}\le \sum_{n=0}^{\infty}\|A_\lambda\|_{\mathcal{L}(X)}^n<\infty.
\end{align*}
The partial sums $S_{\lambda,N}:=\sum_{n=0}^{N}A_\lambda^n$ satisfy
\begin{align*}
(I_X-A_\lambda)S_{\lambda,N}=I_X-A_\lambda^{N+1}.
\end{align*}
Also,
\begin{align*}
S_{\lambda,N}(I_X-A_\lambda)=I_X-A_\lambda^{N+1}.
\end{align*}
Since
\begin{align*}
\|A_\lambda^{N+1}\|_{\mathcal{L}(X)}\le \|A_\lambda\|_{\mathcal{L}(X)}^{N+1}\to 0,
\end{align*}
passing to the operator-norm limit gives
\begin{align*}
(I_X-A_\lambda)S_\lambda=I_X.
\end{align*}
Similarly,
\begin{align*}
S_\lambda(I_X-A_\lambda)=I_X.
\end{align*}
Thus $I_X-A_\lambda$ is invertible in $\mathcal{L}(X)$ and
\begin{align*}
(I_X-A_\lambda)^{-1}=\sum_{n=0}^{\infty}A_\lambda^n.
\end{align*}
[guided]
The point of this step is to isolate the only place where smallness is needed. We have fixed $\lambda_0\in\rho(T)$, so $T-\lambda_0I_X$ is invertible and its inverse is the bounded operator
\begin{align*}
R_0:=R_T(\lambda_0)=(T-\lambda_0 I_X)^{-1}.
\end{align*}
For a nearby scalar $\lambda\in\mathbb C$, define the perturbation operator
\begin{align*}
A_\lambda:=(\lambda-\lambda_0)R_0\in\mathcal{L}(X).
\end{align*}
The hypothesis
\begin{align*}
|\lambda-\lambda_0|<\|R_0\|_{\mathcal{L}(X)}^{-1}
\end{align*}
is exactly the condition that
\begin{align*}
\|A_\lambda\|_{\mathcal{L}(X)}\le |\lambda-\lambda_0|\|R_0\|_{\mathcal{L}(X)}<1.
\end{align*}
We now prove the Neumann-series inversion directly. Since $\mathcal{L}(X)$ is a Banach algebra under composition and the operator norm is submultiplicative, each power $A_\lambda^n$ belongs to $\mathcal{L}(X)$ and satisfies
\begin{align*}
\|A_\lambda^n\|_{\mathcal{L}(X)}\le \|A_\lambda\|_{\mathcal{L}(X)}^n.
\end{align*}
Therefore the series
\begin{align*}
S_\lambda:=\sum_{n=0}^{\infty}A_\lambda^n
\end{align*}
converges absolutely in the operator norm, because it is dominated by the convergent geometric series
\begin{align*}
\sum_{n=0}^{\infty}\|A_\lambda\|_{\mathcal{L}(X)}^n.
\end{align*}
Let $S_{\lambda,N}:=\sum_{n=0}^{N}A_\lambda^n$ denote the $N$th partial sum. Multiplying out the finite geometric sum gives
\begin{align*}
(I_X-A_\lambda)S_{\lambda,N}=I_X-A_\lambda^{N+1}.
\end{align*}
The same computation on the other side gives
\begin{align*}
S_{\lambda,N}(I_X-A_\lambda)=I_X-A_\lambda^{N+1}.
\end{align*}
The error term goes to zero in operator norm, since
\begin{align*}
\|A_\lambda^{N+1}\|_{\mathcal{L}(X)}\le \|A_\lambda\|_{\mathcal{L}(X)}^{N+1}\to 0.
\end{align*}
Taking limits in the two identities gives
\begin{align*}
(I_X-A_\lambda)S_\lambda=I_X.
\end{align*}
Also,
\begin{align*}
S_\lambda(I_X-A_\lambda)=I_X.
\end{align*}
Thus $S_\lambda$ is both a left and right inverse for $I_X-A_\lambda$, so
\begin{align*}
(I_X-A_\lambda)^{-1}=\sum_{n=0}^{\infty}A_\lambda^n.
\end{align*}
[/guided]
[/step]
[step:Factor $T-\lambda I_X$ and obtain the resolvent expansion]
Using $R_0=(T-\lambda_0I_X)^{-1}$, compute
\begin{align*}
(T-\lambda_0I_X)(I_X-(\lambda-\lambda_0)R_0)=T-\lambda_0I_X-(\lambda-\lambda_0)I_X.
\end{align*}
Hence
\begin{align*}
(T-\lambda_0I_X)(I_X-(\lambda-\lambda_0)R_0)=T-\lambda I_X.
\end{align*}
By the previous step, $I_X-(\lambda-\lambda_0)R_0$ is invertible. Since $T-\lambda_0I_X$ is invertible by $\lambda_0\in\rho(T)$, the product $T-\lambda I_X$ is invertible. Therefore $\lambda\in\rho(T)$.
Taking inverses in the product identity gives
\begin{align*}
R_T(\lambda)=(I_X-(\lambda-\lambda_0)R_0)^{-1}R_0.
\end{align*}
Substituting the Neumann series from the previous step,
\begin{align*}
R_T(\lambda)=\left(\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_0^n\right)R_0.
\end{align*}
Right multiplication by the fixed bounded operator $R_0$ is continuous in the operator norm, so
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_0^{n+1}.
\end{align*}
Since $R_0=R_T(\lambda_0)$, this is exactly
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_T(\lambda_0)^{n+1}.
\end{align*}
[guided]
We now connect the Neumann-series inverse to the original resolvent. The operator $R_0$ was defined by
\begin{align*}
R_0=(T-\lambda_0I_X)^{-1}.
\end{align*}
Therefore multiplication in the Banach algebra $\mathcal{L}(X)$ gives
\begin{align*}
(T-\lambda_0I_X)(I_X-(\lambda-\lambda_0)R_0)=T-\lambda_0I_X-(\lambda-\lambda_0)I_X.
\end{align*}
The right-hand side is exactly $T-\lambda I_X$, so
\begin{align*}
(T-\lambda_0I_X)(I_X-(\lambda-\lambda_0)R_0)=T-\lambda I_X.
\end{align*}
The first factor is invertible because $\lambda_0\in\rho(T)$, and the second factor is invertible by the Neumann-series argument from the previous step. Hence their product is invertible in $\mathcal{L}(X)$, so $\lambda\in\rho(T)$.
Taking inverses reverses the order of the product, giving
\begin{align*}
R_T(\lambda)=(I_X-(\lambda-\lambda_0)R_0)^{-1}R_0.
\end{align*}
Substituting the Neumann series
\begin{align*}
(I_X-(\lambda-\lambda_0)R_0)^{-1}=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_0^n
\end{align*}
and using continuity of right multiplication by the fixed operator $R_0$ in the operator norm yields
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_0^{n+1}.
\end{align*}
Since $R_0=R_T(\lambda_0)$, this is the desired expansion
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_T(\lambda_0)^{n+1}.
\end{align*}
[/guided]
[/step]
[step:Conclude holomorphy from the local operator-norm power series]
The previous step proves that for every $\lambda_0\in\rho(T)$, the open disk
\begin{align*}
D_{\lambda_0}:=\{\lambda\in\mathbb C:|\lambda-\lambda_0|<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}\}
\end{align*}
is contained in $\rho(T)$ and that $R_T$ has the operator-norm convergent expansion
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nR_T(\lambda_0)^{n+1}
\end{align*}
for every $\lambda\in D_{\lambda_0}$.
It remains only to identify this local expansion as a holomorphic expansion. For each integer $n\ge 0$, define
\begin{align*}
B_n:=R_T(\lambda_0)^{n+1}\in\mathcal{L}(X).
\end{align*} For every radius $0<r<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}$, the series
\begin{align*}
\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nB_n
\end{align*}
converges uniformly in operator norm on the closed disk $|\lambda-\lambda_0|\le r$, because
\begin{align*}
\|(\lambda-\lambda_0)^nB_n\|_{\mathcal{L}(X)}\le r^n\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{n+1}
\end{align*}
and the resulting geometric majorant converges. The same argument applies to the formally differentiated series
\begin{align*}
\sum_{n=1}^{\infty}n(\lambda-\lambda_0)^{n-1}B_n
\end{align*}
on every smaller closed disk.
We now spell out the termwise differentiation argument. Fix $\mu\in D_{\lambda_0}$ and choose $s>0$ such that
\begin{align*}
|\mu-\lambda_0|<s<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}.
\end{align*}
For all sufficiently small nonzero $h\in\mathbb C$, both $\mu$ and $\mu+h$ lie in the closed disk $|z-\lambda_0|\le s$. For each integer $N\ge 0$, define the finite polynomial map $P_N:\mathbb C\to\mathcal{L}(X)$ by
\begin{align*}
P_N(z):=\sum_{n=0}^{N}(z-\lambda_0)^nB_n.
\end{align*}
This map has derivative
\begin{align*}
P_N'(z)=\sum_{n=1}^{N}n(z-\lambda_0)^{n-1}B_n.
\end{align*}
Define the candidate derivative map $G:\{z\in\mathbb C:|z-\lambda_0|<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}\}\to\mathcal{L}(X)$ by
\begin{align*}
G(z):=\sum_{n=1}^{\infty}n(z-\lambda_0)^{n-1}B_n.
\end{align*}
The [uniform convergence](/page/Uniform%20Convergence) of the differentiated series on $|z-\lambda_0|\le s$ implies that $G$ is continuous there as the operator-norm limit of continuous polynomial maps. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. For each $N\ge 0$, the map $[0,1]\to\mathcal{L}(X)$ given by $t\mapsto P_N(\mu+th)$ is continuously differentiable because $P_N$ is a finite $\mathcal{L}(X)$-valued polynomial. The [fundamental theorem of calculus](/theorems/632) applied to this Banach-valued function gives
\begin{align*}
\frac{P_N(\mu+h)-P_N(\mu)}{h}=\int_0^1 P_N'(\mu+th)\,d\mathcal{L}^1(t).
\end{align*}
Passing to the operator-norm limit as $N\to\infty$ gives
\begin{align*}
\frac{R_T(\mu+h)-R_T(\mu)}{h}=\int_0^1 G(\mu+th)\,d\mathcal{L}^1(t).
\end{align*}
Since $G$ is continuous at $\mu$ and $\mu+th\to\mu$ uniformly for $t\in[0,1]$ as $h\to0$, the integral on the right converges in operator norm to $G(\mu)$. Hence
\begin{align*}
\frac{R_T(\mu+h)-R_T(\mu)}{h}-\sum_{n=1}^{\infty}n(\mu-\lambda_0)^{n-1}B_n
\end{align*}
converges to $0$ in $\mathcal{L}(X)$ as $h\to0$. Thus $R_T$ is complex differentiable at $\mu$. Since $\mu\in D_{\lambda_0}$ and $\lambda_0\in\rho(T)$ were arbitrary, $R_T:\rho(T)\to\mathcal{L}(X)$ is holomorphic in the operator norm.
[guided]
We have already proved the strongest local statement: around each $\lambda_0\in\rho(T)$, the resolvent is represented by a norm-convergent power series. To make the holomorphy conclusion explicit, for each integer $n\ge 0$ define the coefficient
\begin{align*}
B_n:=R_T(\lambda_0)^{n+1}\in\mathcal{L}(X).
\end{align*} Then the expansion from the previous step is
\begin{align*}
R_T(\lambda)=\sum_{n=0}^{\infty}(\lambda-\lambda_0)^nB_n.
\end{align*}
Fix a radius $r$ with
\begin{align*}
0<r<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}.
\end{align*}
For every $\lambda$ satisfying $|\lambda-\lambda_0|\le r$, [submultiplicativity of the operator norm](/theorems/1054) gives
\begin{align*}
\|(\lambda-\lambda_0)^nB_n\|_{\mathcal{L}(X)}\le r^n\|R_T(\lambda_0)^{n+1}\|_{\mathcal{L}(X)}.
\end{align*}
Again by submultiplicativity,
\begin{align*}
\|R_T(\lambda_0)^{n+1}\|_{\mathcal{L}(X)}\le \|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{n+1}.
\end{align*}
Thus
\begin{align*}
\|(\lambda-\lambda_0)^nB_n\|_{\mathcal{L}(X)}\le r^n\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{n+1}.
\end{align*}
The right-hand side is a geometric sequence with ratio
\begin{align*}
r\|R_T(\lambda_0)\|_{\mathcal{L}(X)}<1.
\end{align*}
Hence the Weierstrass majorant argument in the [Banach space](/page/Banach%20Space) $\mathcal{L}(X)$ gives uniform convergence in operator norm on the closed disk $|\lambda-\lambda_0|\le r$.
The differentiated series is
\begin{align*}
\sum_{n=1}^{\infty}n(\lambda-\lambda_0)^{n-1}B_n.
\end{align*}
On any smaller closed disk $|\lambda-\lambda_0|\le s$ with
\begin{align*}
0<s<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1},
\end{align*}
the same estimate gives the majorant
\begin{align*}
n s^{n-1}\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{n+1}.
\end{align*}
This numerical series converges because $s\|R_T(\lambda_0)\|_{\mathcal{L}(X)}<1$. Therefore the differentiated series also converges uniformly in operator norm on smaller closed disks. Now fix $\mu\in D_{\lambda_0}$ and choose $s$ with
\begin{align*}
|\mu-\lambda_0|<s<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}.
\end{align*}
For all sufficiently small nonzero $h\in\mathbb C$, the segment from $\mu$ to $\mu+h$ lies in the closed disk $|z-\lambda_0|\le s$. For every integer $N\ge 0$, define the finite polynomial map $P_N:\mathbb C\to\mathcal{L}(X)$ by
\begin{align*}
P_N(z):=\sum_{n=0}^{N}(z-\lambda_0)^nB_n.
\end{align*}
The finite polynomial derivative is
\begin{align*}
P_N'(z)=\sum_{n=1}^{N}n(z-\lambda_0)^{n-1}B_n.
\end{align*}
Applying the one-variable fundamental theorem of calculus to the Banach-valued map $t\mapsto P_N(\mu+th)$ gives
\begin{align*}
\frac{P_N(\mu+h)-P_N(\mu)}{h}=\int_0^1 P_N'(\mu+th)\,d\mathcal{L}^1(t).
\end{align*}
Let the candidate derivative map $G:\{z\in\mathbb C:|z-\lambda_0|<\|R_T(\lambda_0)\|_{\mathcal{L}(X)}^{-1}\}\to\mathcal{L}(X)$ be defined by
\begin{align*}
G(z):=\sum_{n=1}^{\infty}n(z-\lambda_0)^{n-1}B_n.
\end{align*}
Because the differentiated series converges uniformly in operator norm on $|z-\lambda_0|\le s$, the map $G$ is continuous there as the operator-norm limit of continuous polynomial maps. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $[0,1]$. For each $N\ge 0$, the map $[0,1]\to\mathcal{L}(X)$ given by $t\mapsto P_N(\mu+th)$ is continuously differentiable because $P_N$ is a finite $\mathcal{L}(X)$-valued polynomial. Applying the one-variable fundamental theorem of calculus to this Banach-valued map gives
\begin{align*}
\frac{P_N(\mu+h)-P_N(\mu)}{h}=\int_0^1 P_N'(\mu+th)\,d\mathcal{L}^1(t).
\end{align*}
The uniform convergence of $P_N$ and $P_N'$ on $|z-\lambda_0|\le s$ allows passage to the limit in this identity in the operator norm, giving
\begin{align*}
\frac{R_T(\mu+h)-R_T(\mu)}{h}=\int_0^1 G(\mu+th)\,d\mathcal{L}^1(t).
\end{align*}
Since $G$ is continuous at $\mu$ and $\mu+th\to\mu$ uniformly for $t\in[0,1]$ as $h\to0$, the right-hand side converges in operator norm to
\begin{align*}
G(\mu)=\sum_{n=1}^{\infty}n(\mu-\lambda_0)^{n-1}B_n.
\end{align*}
Thus $R_T$ is complex differentiable at $\mu$, and since $\mu$ was arbitrary in $D_{\lambda_0}$, it is complex differentiable throughout $D_{\lambda_0}$.
Since every point $\lambda_0\in\rho(T)$ has such a disk on which $R_T$ is complex differentiable, the resolvent map
\begin{align*}
R_T:\rho(T)\to\mathcal{L}(X)
\end{align*}
is holomorphic in the operator norm.
[/guided]
[/step]
