Derivative of Complex Power Series/Proof 2

Theorem
Let $\xi \in \C$ be a complex number.

Let $\langle{ a_n}\rangle$ be a sequence in $\C$.

Let $\displaystyle f \left({z}\right) = \sum_{n \mathop =0}^\infty a_n \left({z - \xi}\right)^n$ be a power series in a complex variable $z \in \C$ about $\xi$.

Let $R$ be the radius of convergence of the series defining $f \left({z}\right)$.

Let $\left \vert{z - \xi}\right \vert < R$.

Then:
 * $\displaystyle f' \left({z}\right) = \sum_{n \mathop =1}^\infty n a_n \left({z - \xi}\right)^{n-1}$

Lemma
Define:
 * $\displaystyle g \left({z}\right) = \sum_{n \mathop =1}^\infty n a_n \left({z-\xi}\right)^{n-1}$

Fix an $\epsilon > 0$ satisfying $\epsilon < R-\left\vert{ z-\xi}\right\vert$.

Let:
 * $\displaystyle M = \sum_{n \mathop =2}^\infty \dfrac{n \left({n-1}\right) }{2} \left\vert{ a_n}\right\vert \left({R - \epsilon}\right)^{n-2}$

We use the root test to prove convergence of this series:

The last equality follows from the lemma and:


 * $\displaystyle \limsup_{n \to \infty} \left \vert{ a_n }\right \vert^{1/n} = \dfrac{1}{R}$

Suppose that $\left\vert{ h}\right\vert \le R - \epsilon - \left\vert{ z-\xi}\right\vert$.

It follows by the triangle inequality that $\left\vert{ z - \xi + h }\right\vert \le \left\vert{ z-\xi}\right\vert + \left\vert{ h}\right\vert \le R-\epsilon$.

By the triangle inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:

Letting $h\to 0$ gives $f' \left({z}\right)=g \left({z}\right)$, as desired.

Remark
The proof for real power series is identical.