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.
$$\begin{align*}g(re^{2\pi ix})&=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}\left(re^{2\pi ix}\right)^{n!}\\ &=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}r^{n!}\left(e^{2\pi ia}\right)^{n!/b}\\ &=\sum_{n=1}^b\left(re^{2\pi ix}\right)^{n!}+\sum_{n=b+1}^{\infty}r^{n!}.\end{align*}$$ This shows that $$|g(re^{2\pi ix})|=|\sum_{n=b+1}^{b+q}r^{n!}+\sum_{n=b+q+1}^{\infty}r^{n!}+ \sum_{n=1}^{\infty}(re^{2\pi i x})^{n!}|, q \geq 1.$$ So, I want to show that $$\sum_{n=1}^{b+q}r^{n!}\xrightarrow[r\to 1^-]{}\infty.$$ I want to show $\forall M>0 $, $\exists r_0<1$, such that for some $r_0<r<1$ $|\sum_{n=1}^{b+q}r^{n!}| \geq M$. Let $M>0$ I am getting stuck here with finding such an $r_0$ depending on M. So I appreciate any help with that. Thank you in advance.
You are making things too complicated. Once you get $g(re^{2 \pi i x}) =\sum_{n=1}^{b} (re^{2 \pi i x})^{n!}+\sum_{n=b+1}^{\infty} r^{n!}$ just let $r \to 1$. The first term has a finite limit whereas $\lim_{r \to 1} \sum_{n=b+1}^{\infty} r^{n!} \geq \lim_{r \to 1} \sum_{n=b+1}^{N} r^{n!}=(N-b)$ for each integer $N>b+1$. Hence $lim_{r \to 1} \sum_{n=b+1}^{\infty} r^{n!}=\infty$.