Analytic continuation of $\sum _{n=0}^{\infty} \frac{z^{2^n}} {2^n}$ beyond the unit disc

Question

The function : $\sum _{n=0}^{\infty} \frac{z^{2^n}} {2^n}$ convergs to holomorphic function $f$ on $D_1(0)$ and is continious on $\overline{D_1(0)}$. I need to prove that f can't be exteneded to any domain $\Omega$ such that $\overline{D_1(0)}\subseteq\Omega$.

Any ideas?

Answer 1

If it has an analytical continuation on some domain $\Omega \supset \overline{D}(0,1)$, then $\Omega$ contains some $\overline{D}(0,r)$ for some $r >1$. Using the common properties of analytical functions, this requires the series $\sum_n{2^{-n}r^{2^n}}$ to converge. Well, this series does not.