=== Lemma === {{:Derivative of Complex Power Series/Proof 2/Lemma}}{{qed|lemma}} Define: :$\displaystyle \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$ Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$. Let: :$\displaystyle M = \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}$ 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: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$ 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 = by 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}}