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 $0$?
 A: 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.
The last function you mention does not have a power series expansion in a neighborhood of zero. In fact, it is not continuous at zero, because it blows up if you approach zero along the imaginary axis.
