Prove that $f(z)=\sum_{n=1}^\infty \exp(-n!z)$ has no analytic continuation Prove that $f(z)=\sum_{n=1}^\infty\exp(-n!z)$ has no analytic continuation to any open connected subset of $\mathbb{C}$ that strictly contains $\{z\in\mathbb{C}:\text{Re}z>0\}$.
My proof: If $U$ is an open connected subset of $\mathbb{C}$ that strictly contains $\{z\in\mathbb{C}:\text{Re}z>0\}$, and $f$ can be extended analytically to $U$, then there exists $b\in\mathbb{R}$ such that $bi\in U$. Furthermore, we may assume that $b$ is rational, which means that $n!b$ is an even integer for all $n$ large enough. This means that if $a>0$, then
$$f(a+bi)=C+e^{-n!a}+e^{-(n+1)!a}+e^{-(n+2)!a}+...$$ where $C$ is a complex number. Therefore, we get $|f(a+bi)|\rightarrow \infty$ as $a\rightarrow 0+$ by the monotone convergence theorem. This is means that there isn't even a continuous extension.
My questions: is my proof correct? Also I feel like my proof is too specific to this question, if they change the formula of $f$ to for example $\sum_{n=1}^\infty e^{(-n!+\sin n)z}$ my proof will not work. Are there any better, more general proofs available? Thanks!!
 A: This is a nice question, here is how I'd work.
Start by $g(z)=\exp(-z)$. This maps the right half-plane inside the unit disk. Moreover, the imaginary axis is mapped on the unit circle. Set $h(z)=\sum_{n=1}^\infty z^{n!}$. Then $f(z)=h(g(z))$. If we show that we cannot extend $h$ to any open set properly containing the unit disk, we are done, since $f=h\circ g$.
Suppose that we show that for a point $e^{it}\in\partial\mathbb{D}$, $t\in\mathbb{R}$ it is true that
$$\lim_{r\to1^-}h(re^{it})=\infty$$
Then we cannot extend $h$ to a neighborhood of $e^{it}$, because if we could, say we extended $h$ to $H$, then $\lim_{r\to1^-}|h(re^{it})|=|H(e^{it})|<\infty$.
If we find a dense subset of $\partial\mathbb{D}$ having the property above, then we cannot extend $h$ on any open set that strictly contains the right half plane.
Let $q\in\mathbb{Q}$. Write $q$ as an irreducible fraction $a_q/b_q$ with $b_q\in\mathbb{N}$, $a_q\in\mathbb{Z}$.For any integer $n>b_q$ it is $e^{in!q\pi}=1$, because $n!q=a_qn!/b_q$ and this is an even integer.
We have $$h(re^{iq\pi})=\sum_{n=0}^{b_q}r^{n!}e^{in!q\pi}+\sum_{n=b_q+1}^\infty r^{n!}=A_q(r)+\sum_{n=b_q+1}^\infty r^{n!}.$$
Then
$$|h(re^{iq\pi})|\geq\sum_{n=b_q+1}^\infty r^{n!}-|A_q(r)|,$$
by triangular inequality. Note that $A_q(1)$ is well defined and is some complex number that we won't care about. Now
$$\liminf_{r\to1^-}|h(re^{iq\pi})|\geq\liminf_{r\to1^-}\bigg{(}\sum_{n=b_q+1}^\infty r^{n!}-|A_q(r)|\bigg{)}\geq\sum_{n=b_q+1}^\infty1-|A_q(1)|=\infty,$$
the last inequality coming from Fatou's lemma in measure theory (summation is integration over counting-measure).
Since $\{e^{iq}:q\in\mathbb{Q}\}$ is dense on the unit circle, we are done.
