# Proving that the analytic function defined by the sum goes to infinity for all the rational

Demonstrate the Taylor series $\sum\limits_{n\geq 1}^{\infty}z^{n!}$ has radius of convergece $R =1$, and the analystic function $g(z)=\sum\limits_{n\geq 1}^{\infty}z^{n!}$ has the following property:$$\lim_{r \to 1^-} |g(re^{2x\pi i})| = \infty. \quad \forall x\in \mathbb{Q}$$

$\mathbf{\underline{Solution}}:$

Obviously, we see that $$a_n = \begin{cases} 1;& n = m! \\ 0;& \text{otherwise} \end{cases}$$ So, $$\varlimsup_{n \rightarrow \infty}a_n=\lim_{n \rightarrow \infty} \sup_{k \geq n} a_k=1\\\Rightarrow R=\frac{1}{\varlimsup\limits_{n \rightarrow \infty}a_n}=1,$$ as required.

For the second part that I am struggling with. This is my attempt.

Let $x \in \mathbb{Q}$, then, $x=a/b$ with $a \in \mathbb{Z}, b \in \mathbb{Z}^+$ we have

$$\lim _{r \rightarrow 1^-}|g(re^{2i\pi x})|=\lim _{r \rightarrow 1^-}\left|\sum_{n \geq 1}(re^{2i \pi x})^{n!}\right| \leq \sum_{n \geq 1}|e^{2i \pi x n!}| \cdot \left|\lim _{r \rightarrow 1^-}r^{n!}\right|.$$ Now we see $$\sum_{n \geq 1}|e^{2i \pi x n!}|= \infty,$$ since $|e^{2i \pi x n!}|=1$. But what about $\left|\lim\limits_{r \rightarrow 1^-}r^{n!}\right|$, I see that when $r$ approches $1$ from the left, this limit will be $0$. I am getting stuck here. So I appreciate any help with that. Thank you in advance.

• Note that showing the right-hand side of your inequality goes to $\infty$ is not sufficient, because it's only an upper bound for your sum. – Erick Wong Mar 26 '18 at 0:28
• I think you should have $r\to 1^+$ for the result to hold, otherwise the limit of $0$ is correct. If $r\geq 1$, then $$|g(re^{2i\pi x})|=\sum_{n\geq 1}|r^{n!}|\geq\sum_{n\geq 1}|1^{n!}|\to\infty$$ – Prasun Biswas Mar 26 '18 at 0:36

Note that \begin{align*}g(re^{2\pi ix})&=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}\left(re^{2\pi ix}\right)^{n!}\\ &=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}r^{n!}\left(e^{2\pi ia}\right)^{n!/b}\\ &=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}r^{n!}.\end{align*} This shows that $$|g(re^{2\pi ix})|\geq\sum_{n=b+1}^{\infty}r^{n!}-\left|\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}\right|.$$ So, it is enough to show that $$\sum_{n=b+1}^{\infty}r^{n!}\xrightarrow[r\to 1]{}\infty.$$
• Your index of summation should be $n$, not $i$. – marty cohen Mar 26 '18 at 1:35
• So it is true when $r \rightarrow 1^+$ from the right but I am still not getting if $r \rightarrow 1^-$ – Ahmed Mar 28 '18 at 0:40
• The proof still works if $r\to 1^-$: note that, for any $N\in\mathbb N$, $$\liminf_{r\to 1^-}\sum_{n=b+1}^{\infty}r^{n!}\geq\liminf_{r\to 1^-}\sum_{n=b+1}^{N+b+1}r^{n!}\geq\sum_{n=b+1}^{N+b+1}\liminf_{r\to 1^-}r^{n!}=N,$$ therefore the limit is infinity. – detnvvp Apr 1 '18 at 10:04