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

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?

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.
• Could you please elaborate on the part of "Using the common properties..."? Obviously, I know that the series is a taylor expansion around 0. But maybe, around some $z$ where $|z|>1$, it has another expansion that converges. – user3708158 Jan 22 at 14:45
• I am not sure, for example you have the function $f(z)=\frac {1}{1-z} = \sum _n z^n$ that does not converge on the domain where $|z|>1$. I mean, you could have another expansions around different points in the domain. – user3708158 Jan 22 at 14:48
• That is why I wrote: converges normally on any closed disc that is a subset of the domain. $\overline{D}(0,1)$ (or anything greater than $1$) is not a subset of the domain of $z \longmapsto (1-z)^{-1}$. – Mindlack Jan 22 at 14:50
• Ok, I will change my question: why for points s.t $|z|>1$, I need that series to converge? Why can't I have another expansions, around s.t that do converge? – user3708158 Jan 22 at 14:52