Can an Analytic Function Have Zeroes Approaching the Boundary of Convergence?

[acarchau]

Suppose I have an analytic function, $f$, with radius of convergence $1$. From http://planetmath.org/encyclopedia/ZerosOfAnalyticFunctionsAreIsolated.html" it follows that any $0$ of $f$ in $\{ z : |z| < 1 \}$ is isolated. But is it possible that the zeroes have a limit point on the boundary of convergence? For example, is possible that an analytic function with radius of convergence 1 has zeros exactly at $1-\frac{1}{n}$ for all $n \in Z^{+}$?
[Edit: Added question marks.]

 

[wofsy]

Suppose I have an analytic function, $f$, with radius of convergence $1$. From http://planetmath.org/encyclopedia/ZerosOfAnalyticFunctionsAreIsolated.html" it follows that any $0$ of $f$ in $\{ z : |z| < 1 \}$ is isolated. But is it possible that the zeroes have a limit point on the boundary of convergence. For example, is possible that an analytic function with radius of convergence 1 has zeros exactly at $1-\frac{1}{n}$ for all $n \in Z^{+}$.

Take a look at the Mittag-Leffler Theorem which may be over kill. I do not really understand these issues and would be willing to go through the proofs with you.

 

[Hurkyl]

(It's hard to read that without question marks)


Can you write down an analytic function with infinitely many zeroes?

Can you apply a change of variable to move those zeroes to the unit circle?

 

[acarchau]

Let me try, $$\frac{1}{\Gamma(z)}$$ is entire and has zeroes at $-1,-2, \dots$, so $$\frac{1}{\Gamma(\frac{-1}{1-z})}$$ would have 0's at $1-\frac{1}{n}$ and since the pole closest to 0 is at 1, the radius of convergence is 1.

Does that look right?

[Added later]
$f(\frac{1}{1-z})$ where $f$ is an entire function with zeroes exactly at the positive integers will do in general.