I have read online from two or three sources stating that the following limit is trivial, but I don't see how.
Let $f(z)=\sum_{n=0}^\infty z^{2^n}$, take $r \in \mathbb{R}$ and $\lambda \in \mathbb{C}$ for $\lambda^{2^m}=1$ for $m \in \mathbb{N}$ then: $$\lim_{r \to 1^-} f(r\lambda) = +\infty$$
Would a correct approach to be to express $z$ in terms of $rexp(i\theta)$ and take the limit as $r$ tends to $1$ from the left of $f(r\lambda)$ and somehow make a connection? I see no reason as to why this limit is 'trivial'.
You can look at $\sum_{n=0}^\infty(r\lambda)^{2^n}$ term-wise. The first $m$ terms converge, so you can forget them. The rest of the terms are always positive and they all converge to $1$ as $r\to1^-$. Therefore their sum becomes arbitrarily large.