I would like to show that $\sum_{k=1}^{\infty}\frac{(-1)^{(k-1)}}{k}z^k, z \in \mathbb C$ converges for any $z \ne -1 $ if $ |z|=1$. It would mean (because of Abel's theorem) that $$\sum_{k=1}^{\infty}\frac{(-1)^{(k-1)}}{k}z^k = \log(1+z) , |z| = 1, z \ne 1$$.
I have a hint that the following theorem (Dirichlet criterion) might help:
if $(\sum_{i=1}^{n}a_i)_{n \in \mathbb N}$ is bounded and $\sum_{i=1}^{\infty}|b_{i+1}-b_i| < \infty$ and $b_i \to 0$, then $\sum_{i=1}^{\infty}a_ib_i$ converges.
$(\sum_{k=1}^{n}(-1)^{k-1})_{n \in \mathbb N}$ is always between $0$ and $1$, so bounded. $\frac{z^k}{k} \to 0$ if $|z|=1$. I am not sure how to show that $$\sum_{k=1}^{\infty}|\frac{z^{k+1}}{k+1}-\frac{z^k}{k}|=\sum_{k=1}^{\infty}|\frac{kz-(k+1)}{(k+1)k}|$$ converges.
Solution attempt (inspired by the comment below)
$$\sum_{k=1}^{\infty}| \frac{1}{k+1} - \frac{1}{k}|=\sum_{k=1}^{\infty}|\frac{-1}{k(k+1)}| \le \sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$$ $$\frac{1}{k} \to 0$$
$$|\sum_{k=1}^{n}(-1)^{k-1}z^k|=|z\frac{1-(-z)^{n}}{1+z}| \le \frac{2}{|1+z|}$$.