TITLE: Determine the character of $\sum_{n=1}^{+\infty}{\frac{e^{i\theta n}}{n}}$
QUESTION [5 upvotes]: Determine the character of the following series:
$$\sum_{n=1}^{+\infty}{\frac{e^{i\theta n}}{n}}$$
where $\theta$ is a real parameter.
I try to divide the series with De Moivre' s formula:
$$\sum_{n=1}^{+\infty}{\frac{e^{i\theta n}}{n}}=\sum{\frac{cos(n\theta)}{n}}+i\sum{\frac{sin(n\theta)}{n}}$$
The initial series converges if both these series converge. In particular I have seen that for ${\theta}= \pi+2k\pi$ the series converge and for ${\theta}=2k\pi$ the series diverge.

REPLY [3 votes]: We can divide the study of this series into three cases:
1) θ=0+2kπ, k is an integer
2) θ=2kπ
3) the other cases
then,
1) the series is the harmonic series which does not converge.
2) the series is $$∑(-)^n/n$$ that converges for Leibniz's Rule
3) For the Dirichlet's test the series converges.
In particular in the third point:
$1/n$  is decreasing, strictly positive and approaches to zero at the growth of n.
whit a sum from 1 to A, where A is a positive integer.
$$|∑_1^Ae^{inθ}|$$ $$≤$$ $$∑_1^A|e^{inθ}|$$ $$≤A$$  i.e. it's bounded.
In fact $$|e^{inθ}|$$$$=$$$$1$$ for the rule of the complex field.
So the series for θ≠kπ satisfies the hypothesis of the Dirichlet's Test and then converges.