Let $0 \le p/q < 1$ be a rational number.
Consider holomorphic function $f( z) = \sum z^{n!}$ In unit disc.
In polar form now consider $f(r, 2 \pi p/q) = \sum (r^{n!} )(\exp(2i\pi (p/q) n! ))$.
I want to show that as $r$ tends to $1$ function $f( r, 2 \pi p/q)$ goes to infinity.
I thought of changing the order of limit and sum so that I can use sum $(\exp(2i\pi (p/q) n! ))$ which is indeed a divergent series. But I am unable to make it rigourous.
Can someone please help me with this. Thanks.
Note that for $n \ge q$, we have that $q$ divides $n!$ and thus, $(p/q)n!$ is an integer.
Thus, we have that $\exp(2i\pi(p/q)n!) = 1$ for $n \ge q.$
Note the finite sum $$\sum_{n=0}^{q-1}r^{n!}(\exp(2i\pi (p/q) n! ))$$ converges as $r \to 1$. Thus, we focus only on the sum of the later part, that is,
$$\sum_{n=q}^\infty r^{n!}(\exp(2i\pi (p/q) n! )) = \sum_{n=q}^\infty r^{n!}.$$
Denote the summation on the right by $S(r)$. ($q$ has been fixed from the beginning.)
We wish to show that the above series diverges to $\infty$ as $r\to 1$.
We do this by showing that given any $M \in \Bbb R$, there exists $r_0$ such that for all $r_0 < r < 1$, we have that $S(r) > M$.
Towards this end, let $M$ be given and choose any positive integer $N \ge M$.
Note that $$\lim_{r\to1}r^{(q+2N)!} = 1.$$
Thus, there exists $r_0 \in (0, 1)$ such that $r^{(q+2N)!} > 1/2$ for all $r \in (r_0, 1)$.
Moreover, for any such $r$, we also have $$r^{q!} \ge r^{(q+1)!} \ge \cdots \ge r^{(q + 2N)!} > \dfrac{1}{2}.$$
Thus, we get that
$$\begin{align} S(r) &= \sum_{n=q}^\infty r^{n!}\\~\\ &\ge \sum_{n=q}^{q + 2N} r^{n!}\\~\\ &= r^{q!} + r^{(q+1)!} + \cdots + r^{(q + 2N)!}\\ &> \underbrace{\dfrac{1}{2} + \cdots + \dfrac{1}{2}}_{2N + 1\text{ times}}\\~\\ &= \dfrac{2N+1}{2} = N + \dfrac{1}{2}\\ &> M, \end{align}$$
as desired.