# characterizing the boundary of the convergent region of $f(z)= \sum_{n=1}^{\infty} z^{(1/z)^{n}}$

Let

$$f(z)= \sum_{n=1}^{\infty} z^{(1/z)^{n}}$$

A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:

The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.

The code that was used to generate this is (using mpmath and matplotlib):

from mpmath import *
import pylab
def f(z):
return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])

fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)

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The series converges for $$\left| {{z^{1/z}}} \right| < 1$$ –  Pedro Tamaroff Mar 6 '12 at 21:03
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$ –  Robert Israel Mar 6 '12 at 21:22

$|z^{(1/z)^n}| = \exp(\text{Re}(\log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number, $\log(z) (1/z)^n$ will be in some sector $\{z: -\pi/2+\epsilon \le \arg(z) \le \pi/2 - \epsilon\}$ infinitely often, and since $|\log(z) (1/z)^n| \to \infty$ as $n \to \infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.