Complex power series and radius of convergence

Let $c$ be a non-zero complex number, and consider the power series
\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots.
\end{equation}
By using the Ratio Test, or otherwise, show that the series has radius of convergence $|c|$. By differentiating term by term, show that $S'(z)= \frac{1}{z}$.

I've never done power series in complex analysis, so this is what I've attempted so far:

\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots\\
=\sum_{n=0}^{\infty} \frac{(-1)^n (z-c)^n}{nc}.
\end{equation}
Let $x_{n}=\frac{(-1)^n}{nc}$. Using the Ratio test, we have:
\begin{equation}
lim_{n \rightarrow \infty} \lvert \frac{x_{n+1}(z-c)^{n+1}}{x_{n}(z-c)^n} \rvert = \lvert z-c \rvert lim_{n \rightarrow \infty} \lvert - \frac{n}{n+1} \rvert = -\lvert z-c \rvert <1.
\end{equation}

I'm pretty sure this is wrong somewhere, but I have no idea how to continue to show that the radius of convergence is $|c|$.

Answer

Notice (hint):

First of all:

• When $n$ 'starts' with $0$ than we've got a problem, because we get (dividing by $0$):

$$\frac{(-1)^0(z-c)^0}{0c}$$

• Use the ratio test, to proof that this series converges, when $|c-z|<1$.


• So:

$$\sum_{n=1}^{\infty}\frac{(-1)^n(z-c)^n}{cn}=-\frac{\ln(1+z-c)}{c}\space\space\space\space\space\space\text{when}\space|c-z|<1$$