Radius of Convergence of Derivative of Complex Power Series

Theorem
Let $\xi \in \C$.

For all $z \in \C$, define the power series:
 * $\ds S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$

and:
 * $\ds S' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

Let $R$ be the radius of convergence of $S \paren z$, and let $R'$ be the radius of convergence of $S' \paren z$.

Then $R =R'$.

Proof
Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$.

Then $S' \paren z$ converges absolutely by Existence of Radius of Convergence of Complex Power Series, so:

From the $n$th Root Test, it follows that $S \paren z$ converges absolutely.

Hence, $R \ge R'$.

Suppose that $z \in \C$ with $\cmod {z - \xi} < R$.

Find $z_o \in \C$ such that $\cmod {z - \xi} < \cmod {z_0 - \xi} < R$, so $S \paren {z_0}$ converges absolutely.

From the $n$th Root Test, it follows that $\cmod {a_n \paren {z_0 - \xi}^n }^{1/n} < 1$ for all $n \ge N$ for some $N \in \N$.

Then:

From the $n$th Root Test, it follows that $S' \paren z$ converges absolutely.

Hence, $R' \ge R$, so $R' = R$.