Analytic continuation of $\sum (z/a)^n$

Question

I'm having trouble continuing this function beyond its convergence radius, $R=a$.

$$f(z)=\sum (z/a)^n$$

Given the context (a textbook in complex analysis) I suspect it should have a simple closed-form expression. I've tried differentiating and trying to relate it to the geomtric series, but so far I haven't had any success.

Any hint or idea on how to analytic continuate it?

Thanks in advance!

Answer 1

For $|z/a|<1$ the sum is $\frac{1}{1-(z/a)} = \frac{a}{a-z}$. That is the continuation.

Answer 2

The geometric series $f_{z_0}$ has a series expansion with center $z_0=0$ and radius of convergence $R=|a|$ with $a$ a being a simple pole. \begin{align*} f_{z_0}(z)=\sum_{n=0}^\infty\left(\frac{z}{a}\right)^n=\frac{a}{a-z} \end{align*}

When we consider a series expansion around another point $z_1$ we know that the radius of convergence is the distance from $z_1$ to the nearest singularity which is $a$ in this case.

In order to obtain an analytic continuation of $f_{z_0}(z)=\frac{a}{a-z}$ we can select a point $z_1$ within the disc $D_{z_0}$ of convergence of the series expansion of $f_{z_0}$ around $z_0=0$. This way we obtain a function $f_{z_1}$

\begin{align*} f_{z_1}(z)&=\frac{a}{a-z}\\ &=\frac{a}{(a-z_1)-(z-z_1)}\\ &=\frac{a}{a-z_1}\cdot\frac{1}{1-\frac{z-z_1}{a-z_1}}\\ &=\frac{a}{a-z_1}\sum_{n=0}^\infty\left(\frac{z-z_1}{a-z_1}\right)^n \end{align*}

This function converges within the disc $D_{z_1}$ with center $z_1$ and radius $|z_1-a|$.

Depending on the choice of $z_1$ the disc $D_{z_1}$ may contain points which are not inside $D_{z_0}$. This can be the starting point for creating a germ of $f_{z_0}$.