Theorem: Derivative of Complex Power Series

Let $\xi \in \C$ be a complex number.
Let $\sequence {a_n}$ be a sequence in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a power series in a complex variable $z \in \C$ about $\xi$.
Let $R$ be the radius of convergence of the series defining $\map f z$.
Let $\cmod {z - \xi} < R$.
Then $f$ is complex-differentiable and its derivative is:
:$\ds \map {f'} z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

Proof:
=== Lemma ===
{{:Derivative of Complex Power Series/Proof 2/Lemma}}{{qed|lemma}}
Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Let:
:$\ds M = \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}$
We use the Root Test to prove convergence of this series:
{{begin-eqn}}
{{eqn | l = \limsup_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2} }^{1/n}
      | r = \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \lim_{n \mathop \to \infty} \paren {\paren {R - \epsilon}^{n - 2} }^{1/n} \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}
}}
{{eqn | r = \dfrac {R - \epsilon} R
}}
{{end-eqn}}
The last equality follows from the lemma and:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$
Suppose that $\size h \le R - \epsilon - \cmod {z - \xi}$.
It follows by the Triangle Inequality that:
:$\cmod {z - \xi + h} \le \cmod {z - \xi} + \size h \le R - \epsilon$
By the Triangle Inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:
{{begin-eqn}}
{{eqn | l = \cmod {\dfrac {\map f {z + h} - \map f z} h - \map g z}
      | r = \cmod {\sum_{n \mathop = 2}^\infty a_n \paren {\dfrac {\paren {z - \xi + h}^n - \paren {z - \xi}^n} h - n \paren {z - \xi}^{n - 1} } }
}}
{{eqn | o = \le
      | r = \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\dfrac {\paren {z - \xi + h}^n - \paren {z - \xi}^n} h - n \paren {z - \xi}^{n - 1} }
      | c = Triangle Inequality
}}
{{eqn | r = \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\paren {\sum_{u \mathop = 0}^{n - 1} \paren {z - \xi + h}^u \paren {z - \xi}^{n - u - 1} } - n \paren {z - \xi}^{n - 1} }
      | c = Difference of Two Powers
}}
{{eqn | r = \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\sum_{u \mathop = 0}^{n - 1} \paren {\paren {z - \xi + h}^u \paren {z - \xi}^{n - u - 1} - \paren {z - \xi}^{n - 1} } }
}}
{{eqn | r = \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\sum_{u \mathop = 0}^{n - 1} \paren {z - \xi}^{n - u - 1} \paren {\paren {z - \xi + h}^u - \paren {z - \xi}^u } }
}}
{{eqn | r = \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {h \sum_{u \mathop = 0}^{n - 1} \paren {z - \xi}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \paren {z - \xi + h}^v \paren {z - \xi}^{u - v - 1} }
      | c = Difference of Two Powers
}}
{{eqn | o = \le
      | r = \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{u \mathop = 0}^{n - 1} \cmod {z - \xi}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \cmod {z - \xi + h}^v \cmod {z - \xi}^{u - v - 1}
      | c = Triangle Inequality
}}
{{eqn | o = \le
      | r = \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{u \mathop = 0}^{n - 1} \paren {R - \epsilon}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \paren {R - \epsilon}^v \paren {R - \epsilon}^{u - v - 1}
      | c = by assumption and $\cmod {z - \xi + h} \le R - \epsilon$, which was proven earlier
}}
{{eqn | r = \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^{n - 2} \sum_{u \mathop = 0}^{n - 1} \sum_{v \mathop = 0}^{u - 1} 1
}}
{{eqn | r = \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^{n - 2} \sum_{u \mathop = 0}^{n - 1} u
}}
{{eqn | r = \size h \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}
      | c = Closed Form for Triangular Numbers
}}
{{eqn |r = M \size h
}}
{{end-eqn}}
Letting $h \to 0$ we see that $\map {f'} z$ exists and $\map {f'} z = \map g z$, as desired.
{{qed}}