Derivative of Complex Power Series/Proof 2

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:

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:

Letting $h \to 0$ we see that $\map {f'} z$ exists and $\map {f'} z = \map g z$, as desired.

Also see
The proof for real power series is identical.