Convergence of $\sum \frac{(-1)^{n+1}}{n}z^n$ at $|z|=1$ I know that the power series $\sum \frac{(-1)^{n+1}}{n}z^n$ converges for $|z| \lt 1$ but I have been trying to determine what happens on $|z|=1$
Clearly the series converges at $z=1$ and diverges at $z=-1$.
I thought that a good approach would be to write $z^n = \cos n\theta + i\sin n\theta$ and use Dirichlet's test for the two real series $\sum \frac{(-1)^{n+1}}{n} \cos n\theta, \space \sum \frac{(-1)^{n+1}}{n} \sin n\theta$ but I can't really use it here because $\frac{(-1)^{n+1}}{n}$ is not monotonic.
Can someone give a hint with the right direction?
Let $z=e^{it}$, then the series becomes

$$-\sum_{n=1}^\infty\frac{(-1)^ne^{int}}{n}$$

Let $a_n=(-1)^ne^{int}$, then for $t\neq\pi$:
$$\left|\sum_{n=1}^Na_n\right|=\left|\frac{-e^{it}(1-(-1)^Ne^{iNt})}{1+e^{it}}\right|\leq\left|\frac{2}{1+e^{it}}\right|$$

Then apply Dirichlet test to $a_n$ and $b_n=\frac{1}{n}$, we conclude that $\sum_{n=1}^\infty a_nb_n$ converges, for $z=e^{it}, t\neq\pi$.

If $t=\pi$, then it becomes the harmonic series, which diverges.
Hint:

For $\lvert z\rvert = 1$, consider the series representation of 
$$
\log(z) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(z-1)^n.
$$ 
What happens for $z$ on the unit circle?