How to prove a power series can be analytically continued?

Question

Consider the complex function $f(z)=\sum_{n=1}^{\infty} \frac{(-1)^n}{2n+1}z^{2n+1} \, .$

(i) Determine its domain.

(ii) Let $\Gamma = \{iy:y\in\mathbb{R}, |y| \geq 1 \}$, show that there exists an analytic continuation of $f$ to $\mathbb{C}-\Gamma$.

(iii) (optional) Discuss the existence of other maximal analytical continuations of $f$ to domains of the Riemann sphere $\mathbb{C} \cup \{\infty\}$ of the form $D= (\mathbb{C} \cup \{\infty\})-\Gamma',$ where $\Gamma'$ is a compact set of $\mathbb{C}$.

For the first part: I think the radius of convergence is $1$. And I can see it converges when $z=1$, but I don't know about the rest of the border $S^1$. However, the real problem is point (ii). Is there a general way to tackle this kind of problem?

Answer 1

HINT:

For $|z|<1$, we have

$$f'(z)=\sum_{n=1}^{\infty} (-z^2)^n \tag 1$$

Now, find $f(z)$ by evaluating the geometric series in $(1)$ and integrating the result. Finally, note that the function $f(z)$ extended beyond the unit circle has branch points at $\pm i$.

Answer 2

I doubt very much that there is a very general way to determine when a power series has an analytic continuation outside its radius $R$ of convergence. You're unlikely to be able to find a domain for a maximal analytic continuation unless you can find a closed form for the function (or for the difference between the function and some entire function).

One thing you can sometimes do is to identify a finite number of points of the circle of radius $R$ where the series "goes bad", and see whether subtracting a suitable function with poles or branch points at those points leaves you with a series having greater radius of convergence.

There are also some sufficient conditions for the function to have a natural boundary on the circle of convergence, notably the Ostrowski–Hadamard gap theorem.