I know I could use Dirichlet's test, but I am wondering if the Taylor series of $- \ln (1-z)$ can be used in some way to prove it for $\|z\|=1$, $z \neq 1$. I know the convergence radius is 1 so it is not straightforward.
I tried to change the center of the Taylor serie and used $y \in \mathbb{C}$ instead of zero and I obtained that if $f(z):= - \ln(1-z)$ \begin{align*} f(z+y-zy) &=\sum_{n=1}^{\infty}\frac{f^{n)}(y)}{n!}((z+y-zy)-y)^n = \\ &=\sum_{n=1}^{\infty}\frac{(n-1)!}{(1-y)^n}\frac{(z-zy)^n}{n!}=\sum_{n=1}^{\infty}\frac{z^n}{n} \end{align*} And for $\|z\|=1$, $z \neq 1$ the logarithm is finite but for $z = 1$ it is infinity but I think this is not a correct argument because I think that the Taylor serie is equal to the logarithm only if $(z+y-zy)$ is inside of a ball with $y$ as center and that ball is in an open domain where the logarithm is analytic, in this case this is that the ball doesn't intersect the points where the logarithm doesn't exists.
How could I use the Taylor serie of the logarithm to prove the convergence of the series $\sum_{n=1}^{\infty} \frac{z^n}{n}$ for $z \in \mathbb{C}$ such that $\|z\|=1$ but $z \neq 1$?