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
$$x\in(-\infty,1)~.$$
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?
Thanks in advance,
Alex