# Radius of Convergence of Derivative of Complex Power Series

## Theorem

Let $\xi \in \C$.

For all $z \in \C$, define the power series:

$\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$

and:

$\displaystyle S' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

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:

 $\displaystyle 1$ $\ge$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1 / n}$ $\quad$ by $n$th Root Test $\quad$ $\displaystyle$ $>$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n }^{1/n}$ $\quad$ as $\cmod {n \paren {z - \xi}^{n - 1} } > \cmod { \paren {z - \xi}^n }$ for all $n > \cmod {z - \xi}$ $\quad$

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

Hence, $R \ge R'$.

$\Box$

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:

 $\displaystyle 1$ $=$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod n^{1/n}$ $\quad$ by Limit of Integer to Reciprocal Power $\quad$ $\displaystyle$ $>$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod n^{1/n} \cmod {\frac {z - \xi} {z_0 - \xi} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n}$ $\quad$ as $\displaystyle \lim_{n \mathop \to \infty} \paren {\paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n} = 1$ $\quad$ $\displaystyle$ $\ge$ $\displaystyle \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} a_n \paren {z_0 - \xi}^{n - 1} }^{1/n}$ $\quad$ from the $n$th Root Test, as remarked above $\quad$ $\displaystyle$ $=$ $\displaystyle \limsup_{n \to \mathop \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1/n}$ $\quad$ $\quad$

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

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

$\blacksquare$