Discontinuous power series at radius 1

Question

How to show that the power series $$ f(z) = \sum \limits_{n=1}^{\infty}\frac{z^{3^{n}}-z^{2\cdot 3^{n}}}{n} $$ is discontinuous at $z=1$?
I already know the series diverges to $-\infty$ for $z=e^{i\pi / 3^{k}}$ and $k \in \mathbb{N}$, and I tried showing that $f(z)<-M$ for any positive number $M$, if $z$ is sufficiently close to $e^{i\pi / 3^{k}}$, but I didn't succeed in proving it in any rigorous manner.

Answer 1

For $z=e^{\pi i\cdot 3^{-k}}$ and $n\ge k$, $\frac{z^{3^{n}}-z^{2\cdot 3^{n}}}{n}=\frac{e^{\pi i\cdot 3^{n-k}}-e^{2\pi i\cdot 3^{n-k}}}{n}=-\frac{2}{n}$, then $f(z)=C-\sum\limits_{n\ge k}\frac{2}{n}=-\infty$.

This power series $ \sum _{n>0}{\frac {z^{3^{n}}-z^{2\cdot 3^{n}}}{n}}$ converges to $0$ at $z=1$, but is unbounded at any point of the form $ e^{\pi i/3^{n}}$, so the value at $z=1$ is not the limit as $z=e^{\pi i/3^{n}}$ tends to 1 in the whole open disk. You can see it enter link description here.