$\sum\limits_{k=1}^{\infty}\frac{z^k}k$ for $|z|=1$

The sum $$\sum\limits_{k=1}^{\infty}\frac{z^k}k$$ is given and I want to find out for which $z\in\mathbb C$ the series converges. By applying the ratio test or calculating the radius of convergence, we learn that the series converges for $|z|<1$ and diverges for $|z|>1$.

But what about the case $|z|=1$? If $z=1$, for example, the harmonic series does not converge, however, for $z=-1$ the series converges (alternating series test), so the convergence behaviour isn't the same for all $|z|=1$.

I've already tried writing $z$ as $a+\rm i\sqrt{1-a^2}$, but could not get anything helpful from that. How should I proceed?

• The series converges for all $z\ne 1$. – Mark Viola Dec 1 '16 at 16:35
• You'll want to use the Dirichlet test here – Omnomnomnom Dec 1 '16 at 16:36
• WLOG $z=e^{2ix}$ $$\dfrac{z^k}k=\dfrac{(e^{2ix})^k}k$$ $$S=\sum_{k=1}^\infty\dfrac{z^k}k=-\ln(1-e^{2ix})=-\ln\{-e^{ix}(e^{ix}-e^{-ix})\}$$ For $\pi>x>0,$ $$S=-\ln(-i)-ix+\ln(2\sin x)=-i\pi-\dfrac{i\pi}2-ix+\ln(2\sin x)$$ – lab bhattacharjee Dec 1 '16 at 16:40

Hint. Write $z = e^{it}$, where $t$ is not an integral multiple of $2\pi$ (so that $z \neq 1$). So we consider the series $\sum\limits_{k = 1}^\infty \frac{e^{ikt}}{k}$. Show that the partial sums $\sum\limits_{k = 1}^N e^{ikt}$ are bounded by $\lvert \csc(t/2)\rvert$. Conclude that the series converges by Dirichlet's test.