# Analytic continuation of $z-z^2+z^3-…$

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

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

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

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!

• This is a geometric series, so you should be able to compute its sum explicitly. (Then you can see how to do the analytic continuation.) – mrf Jan 4 '16 at 14:52
• In this case, you can write down the closed form of $f$, and that immediately gives you an analytic continuation to almost the whole plane. – Daniel Fischer Jan 4 '16 at 14:53

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.
• Thanks! But what happens if $|z|>1$? – G. Schiele Jan 5 '16 at 14:07
• Yes, I mean what happens to the original function. If $|z|<1$ we can continue it through the closed form, but we haven't said anything about the $|z|>1$ case. – G. Schiele Jan 5 '16 at 15:51
• @G.Schiele The original function is only defined for $|z| <1$... To keep things simple, let us use $f$ for the original function and $g$ for the closed form. Then $f(z)$ is only defined for $|z| <1$, and nowhere else (as the series is divergent for all $|z| \geq 1$). Meanwhile $g(z)$ is defined on $\mathbb C \backslash \{ -1 \}$ and, on the domain of $f$ we have $$f=g$$ – N. S. Jan 5 '16 at 16:16
• There is nothing else you can say about $f$ on $|z| \geq 1$ other than $f$ is NOT DEFINED. – N. S. Jan 5 '16 at 16:17
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.