Let's say that $S_n={\lvert \sum_{k=0}^{n}z^k \rvert }$
My intuition is that since $$\sum_{k=0}^{n}z^k = \sum_{k=0}^{n}{e^{i(k\theta)}}=\frac{\sin{\frac{(n+1)}{2}\theta}}{\sin{\frac{\theta}{2}}}{e}^{i\frac{n\theta}{2}}$$, ←(a)
the sequence $\{S_n\}$ "somewhat" looks like a periodic function where $0\leq S_n \leq \frac{1}{\lvert \sin{\frac{\theta}{2}} \rvert} $. ←(b)
There always exists some positive $r$ such that $r\leq S_n \leq \frac{1}{\lvert \sin{\frac{\theta}{2}} \rvert} $ for infinite number of $n$'s. ←(c)
Now I will construct a subsequence $\{s_{n}\}$ from $\{S_n\}$ which satisfies the inequality above, so that $\lim_{n\rightarrow\infty}{s_n}^{1/n} = \lim_{n\rightarrow\infty}{r}^{1/n}= \lim_{n\rightarrow\infty}{{\frac{1}{\lvert \sin{\frac{\theta}{2}} \rvert}}}^{1/n} = 1$. ←(d)
Therefore, the upper limit of $\{S_n\}$ is $1$. ←(e)
I feel like the each step of the intuition is pretty clear, but it lacks precision when moving on to the next step, especially from (b) to (c). What should I add more?
The limit $\lim_{n\to\infty}S_n\le(\sum|z^k|)^{1/n}=\lim n^{1/n}=1$, by the triangle inequality.
On the other hand, the limit is attained at $z=1$.
Thus the limsup is $1$.