# Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion?

Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion? I'm referring to analytic fuctions of course (i.e. those with power series expansions but not necessarily analytic over the entire complex plane).

• You say you're referring to analytic functions which are not analytic. Perhaps you mean something else? – Sharkos May 22 '13 at 8:52
• @Sharkos Sorry I mean functions that are analytic at least somewhere, i.e. not necessarily entire functions. – Ryan May 22 '13 at 8:56

The question is a little confusing, but how about $f(z) = \dfrac{1}{1-z}$.
The power series of $f$ around $z=0$ is $\sum_{n=0}^\infty z^n$ and this converges on $|z|<1$. The function, however, is analytic on $\mathbb{C}\setminus \{1\}$.
• From the comments above, now you can see why the phrasing is confusing. You need to think of $f$ and the power series of $f$ as separate entities. If you start with an analytic function on $\Omega$, you can compute its power series around any point $a\in\Omega$, but this series will only converge on a disc, not necessarily on the whole of $\Omega$. On the other hand, if you start with a power series, converging on some disc, it may very well have an analytic continuation to a larger domain. – mrf May 22 '13 at 9:08