# Why do complex functions have a finite radius of convergence?

Say we have a function $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$ with radius of convergence $R>0$. Why is the radius of convergence only $R$? Can we conclude that there must be a pole, branch cut or discontinuity for some $z_0$ with $|z_0|=R$? What does that mean for functions like
$$f(z)=\begin{cases} 0 & \text{for z=0} \\\ e^{-\frac{1}{z^2}} & \text{for z \neq 0} \end{cases}$$
that have a radius of convergence $0$?

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@KH: Please rectify your question. I am not able to see it clearly. – anonymous Aug 12 '10 at 12:55
@Chandru1: Fixed. Thanks for catching it. – Larry Wang Aug 12 '10 at 13:08

If the radius of convergence is $R$, that means there is a singular point on the circle $|z| = R$. In other words, there is a point $\xi$ on the circle of radius $R$ such that the function cannot be extended via "analytic continuation" in a neighborhood of $\xi$. This is a straightforward application of compactness of the circle and can be found in books on complex analysis, e.g. Rudin's.
However, it does not mean that there is a pole, branch cut, or discontinuity, though those would cause singular values. Indeed, a "pole" on the boundary would only make sense if you can analytically continue the power series to some proper domain containing the disk $D_R(0)$, and this is generally impossible. For instance, the power series $\sum z^{2^j}$ cannot be continued in any way outside the unit disk, because it is unbounded along any ray whose angle is a dyadic fraction. The unit circle is its natural boundary, though it does not make sense to say that the function has a branch point or pole there. (More generally, one can show that given any domain in the plane, there is a holomorphic function in that domain which cannot be extended any further, essentially using variations on the same theme.)
The function $\sum_j \frac{z^j}{j^2}$, incidentally, is continuous on the closed unit disk, but even though there is a singular point there. So continuity may happen at singular points.
Could you please tell me why the function $\sum_ \frac{z^j}{j^2}$ has singular point on closed unit disk?at which point it is? – La Belle Noiseuse Apr 27 '13 at 7:50
@Tsotsi: There's a singular point at $z = 1$; this is a special case of a theorem of Pringsheim. See books.google.com/… – Akhil Mathew Apr 28 '13 at 1:44