Analytic continuation of $z-z^2+z^3-...$

Question

I'm having trouble with the concept of analytic continuation of power series beyond the radius of convergence. For example for:

$$f(z)=z-z^2+z^3-z^4+\cdots=\sum_{n=0}^\infty(-1)^nz^{n+1}$$

I get the convergence radius

$$R=\frac{1}{\lim\sup\sqrt[n]{|(-1)^n|}}=1$$

I've seen the proof that there's at least a singular point on the frontier, but I'm not sure how to show to what extent $f$ can be analytically continued.

Any explanation or hint would be appreciated!

Answer 1

Using the formula for the Geometric series you get $$f(z)=z-z^2+z^3-z^4+...=z(1-z+z^2-z^3+....)=z \cdot \frac{1}{1-(-z)}=\frac{z}{1+z}$$ for $|z| <1$.

Now, $\frac{z}{1+z}$ is analytic on $\mathbb C \backslash \{-1 \}$ and agrees with your power series in your disk of convergence. This is what we mean by analytic continuation.

Answer 2

Multiplying $$f(z)=z-z^2+z^3-z^4+\cdots$$ by $z$ gives $$zf(z)=z^2-z^3+z^4-z^5+\cdots.$$ Adding the two gives $$zf(z)+f(z)=z,$$ so that $$f(z)(z+1)=z,$$ from which we obtain $$f(z)=\frac{z}{z+1}\tag{$\mid z\!\!\mid<1$}.$$ But this expression is analytic on all $\mathbb{C}\setminus\{-1\}$ and agrees with $f(z)$ for $|z|<1$ so the right hand side is the analytic continuation of $f(z)$ to the punctured complex plane.