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Suppose $\theta$ is not an integer multiple of $\pi$.

The series $ \left | \sum e^{i n \theta} \right |$ is bounded above by $\frac{1}{|\sin \theta|}$ and, as $\left ( \frac{1}{n} \right ) $ is decreasing and tends to $0$, the series $$\sum \frac{e^{i n \theta}}{n}$$ converges, according to Dirichlet's test.

As I saw in an answer of this question, the sum is equal to $- \ln (1 - e^{i \theta})$, however, I could not think about any way which would lead to this result.

Does someone know or has any references for the derivation of this result?

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El Cid
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    Do you know how to calculate the Taylor series for the logarithm $\ln(1-x) = -\sum \frac{x^n}{n}$? If so then just plug in $x = e^{i\theta}$ (justified by Abel's Theorem since you have shown that the series converge for $\theta \not = 0$) to get the result. – Winther May 31 '15 at 22:56
  • @ElCid I provided a solution methodology that applies to a wide class of series of this type. Let me know how I can improve my answer. I just want to give the best answer I can give you. – Mark Viola Jun 01 '15 at 00:48

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Here is a "trick" that we can use for a broad class of problems of this type.

Let's examine the function $f(\lambda)$ with series representation

$$f(\lambda)=\sum_{n=1}^{\infty}\frac{\lambda^ne^{in\theta}}{n}$$

where $|\lambda|<1$. Now, we can differentiate term by term to reveal that

$$\begin{align} f'(\lambda)&=\frac{1}{\lambda}\sum_{n=1}^{\infty}\lambda^ne^{in\theta}\\\\ &=\frac{e^{i\theta}}{1-\lambda e^{i\theta}} \end{align}$$

whereupon integrating, enforcing $f(\lambda =0)=0$, and letting $\lambda \to 1$, we find that for $0<\theta <2\pi$

$$\begin{align} f(1)&=-\log(1- e^{i\theta})\\\\ &=-\log 2\sin(\phi/2)+i\left(\frac{\pi-\phi}{2}\right) \end{align}$$

Mark Viola
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