I am currently trying to get a better grasp of the concept of analytic continuation and so I am working through this example:

Let $$f:(-1,1)\to\mathbb{R}~,~x\mapsto\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$

Then we can use Taylor expansion to get:

$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} =\sum_{n=0}^\infty \frac{n!(x-x_0)^n}{n!(1-x_0)^{n+1}} =\frac{1}{1-x_0}\sum_{n=0}^\infty \left(\frac{x-x_0}{1-x_0}\right)^n$$

Now, what we have is another geometric series, which converges for $$\left|\frac{x-x_0}{1-x_0}\right|<1$$

and in fact, the whole thing even converges to $$\frac{1}{1-x}$$ so all is nice here. The question is, for which $x$ does it converge? We get:

$$\left|\frac{x-x_0}{1-x_0}\right|<1\Leftrightarrow |x-x_0|<1-x_0~,$$ since $1-x_0 > 0$. To maximize the RHS we let $x_0\to-1$ and we get

$$|x+1|<2\Leftrightarrow -3<x<1~.$$

Now we can repeat this procedure as many times as we like and always push the negative boundary but never the positive so that we get


Finally to get to my question: Can you continue $f$ in the positive direction, using Taylor expansion or do you need another method for that? Does anyone know a reasonably simple way how to do this?

Thanks in advance,

  • $\begingroup$ You have to use analytic continuation with complex numbers, otherwise you will always find a singularity at $x=1$. $\endgroup$
    – Crostul
    Sep 18, 2019 at 8:16
  • 1
    $\begingroup$ @Crostful You'll get a singularity at $z=1$ no matter what you do. $\endgroup$ Sep 18, 2019 at 8:36

2 Answers 2


The analytic continuation is performed in the complex plane, where the right side of $$\sum_{n\in\mathbb{N}} z^n = \frac{1}{1-z}$$ shows its unique pole at $z=1$. By the Uniqueness of Analytic Continuation for complex functions, $$\forall z\in\mathbb{C}\setminus\{1\} \\ f(z)=\frac{1}{1-z}$$ must be the analytic continuation of the geometric series, since it defines the same correspondence rule than $\sum_{n\in\mathbb{N}} z^n$ at the open unit $z$-complex disk.

This power series expansion turn into a Fourier series expansion if we simply set some $t\in\mathbb{R}$ such that $$z\to e^{it}.$$


As @Crostul said, you can't expand it to positive numbers after $1$, because your function isn't continuous at $x=1$. But you can consider $g(x) = \frac{1}{1+x} = \sum_{k=0}^{+\infty} (-1)^kx^k$ to get a function like the previous one that its boundary can extend to positive numbers to $+\infty$. Although from negative side can't extend like the previous one for similar reason.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .