The series $\sum \frac{z^n}{n}$ converge $\forall z\in \mathbb{C}, |z|=1$ but not $z=1$

Question

I see the answers using Abel's test, That is quite acceptable, but in my text book the hint is gives as "$\sum_{k=1}^n \cos(k\theta)~\sum_{k=1}^n\sin(k\theta) $ are bounded when $0<\theta<2\pi$"

I don't even understand whether the given hint is true and how to use it. Can someone help me?

Answer 1

I don't see how to use Abel's test here. It is natural to use Dirichlet's test: if $N\in\mathbb N$, is $\lvert z\rvert=1$ and if $z\neq1$, then$$\left\lvert\sum_{n=1}^Nz^n\right\rvert=\left\lvert\frac{z-z^{N+1}}{1-z}\right\rvert\leqslant\frac2{\lvert z-1\rvert}.$$So, since $\left(\frac1n\right)_{n\in\mathbb N}$ decreases and converges to $0$…