For what $k$ does the following series converge: $$\sum_{n=1}^{\infty} \frac{e^{i kn}}{n}$$
According to the ratio test it converges when $L=\left|e^{ik}\right|<1$
But $L=\left|e^{ik}\right|=\left|\cos(ik)+i \sin(ik)\right|=\sqrt{\cos^2(ik)+i \sin^2(ik)}=1$ So, does it mean that the series diverges? Wolframalpha says, that the series converges when $e^{im(k)} \ge 1$ and $e^{ik} \ne 1$. How to get this result?
Note that $\left|e^{ik}\right|=e^{\operatorname{Re}(ik)}=e^{-\operatorname{Im}k}$. So, if $\operatorname{Im}k>0$ your series converges and if $\operatorname{Im}k<0$ it diverges, by the ratio test.
If $\operatorname{Im}k=0$, that is, if $k\in\Bbb R$, it still converges, unless $k=0$ (it is clear that your series diverges if $k=0$, since then it's the harmonic series). That follows from Dirichlet's test: the partial sums of the series $\sum_{n=1}^\infty e^{ikn}$ are bounded and the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is a real monotonic sequence which converges to $0$.