Limit of holomorphic function 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. 
 A: 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.
A: I'll just add that
$$\lim_{r\to 1^-}\sum_{n=q}^{\infty}r^{n!}  =\sum_{n=q}^{\infty}\lim_{r\to 1^-}r^{n!} = \sum_{n=q}^{\infty} 1 =\infty$$
follows quickly from the monotone convergence theorem.
