# Conditional convergence of a power series on the $z \in \mathbb{C}$ for which $|z|=\rho$ [duplicate]

Consider the following complex power series $$\sum_{n \geq 1} \frac{z^n}{n} \,\,\,\,\,\,\, z \in \mathbb{C}$$

It surely converges conditionally for $z=-1$ (for alternating series test) and for $z=1$ it diverges (it is the harmonic series).

My question is: how can one show that the power series converges conditionally for any $z \in \mathbb{C}$ such that $|z|=1$ (except for $z=1$)?

## marked as duplicate by Martin R, Daniel Fischer♦ complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 12 '16 at 19:30

We can use the Dirichlet's test to prove the desired result: with $a_n=\frac1n$ and $b_n=e^{in\theta}$ we verify that
• $(a_n)$ is decreasing to 0
• $\sum_{n=1}^N b_n$ is bounded
so the series $\sum a_nb_n$ is convergent.