# Analytic continuation of geometric series using Taylor expansion

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

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

$$x\in(-\infty,1)~.$$

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?

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