Does the Series $\sum_{\substack{k=-\infty\\k\neq0}}^{\infty}\frac{e^{\frac{-\pi ik}{5}} - 1}{k}$ Converge?

Question

I'm trying to prove that this series converges, although I'm not entirely convinced that it does:

$$\sum_{\substack{k=-\infty\\k\neq0}}^{\infty}\frac{e^{\frac{-\pi ik}{5}} - 1}{k}$$

This is a two-sided infinite series of Fourier coefficients for a real valued 1-periodic function on $\mathbb{R}$, if that makes any difference. I haven't been able to find a way, does anyone have any ideas? Thanks.

Edit: Maybe the best way would be to somehow adapt the fact that the sum of the nth roots of unity is equal to zero..

Answer 1

The sum does not converge absolutely as $$\sum_{\substack{k=-\infty\\k\neq0}}^{\infty}\left|\frac{e^{\frac{-\pi ik}{5}} - 1}{k}\right| = 4 \sum_{k=1}^\infty \frac{|\sin (k\pi/10)|}k \geq \frac{4}{5}\sum_{j=0}^\infty \frac{1}{2j+1} = \infty.$$ For the inequality, I have taken only the terms with $k=5(2j+1)$.

Maybe you are interested in another notion of convergence; summing the terms in the order $k=1,-1,2,-2, \dots$ leads to a finite result...

Edit:

As it turns out the OP is also interested in the sum $$S=\sum_{k=1}^{\infty}\frac{e^{\frac{-\pi ik}{5}} - 1}{k}.$$ Let us look at the real part of this sum. We have $$\mathop{\rm Re} S = \sum_{k=1}^{\infty}\frac{\cos(\pi k/5) - 1}{k}.$$ All terms in the sum are negative, so we can find a upper bound by only taking the terms with $k=5(2j+1)$, $j\in\mathbb{N}_0$. Thus, $$\mathop{\rm Re} S \leq - \sum_{j=0}^\infty \frac{2}{2j+1} =-\infty$$ and the series thus not converge.

Answer 2

The two series $$ \sum_{k=-\infty}^{-1}\frac{e^{\frac{-\pi ik}{5}} - 1}{k}, \qquad \sum_{k=1}^{\infty}\frac{e^{\frac{-\pi ik}{5}} - 1}{k} $$ both diverge, but the "principal value" $$ \lim_{K \to \infty} \;\left(\sum_{k=-K}^{-1}\frac{e^{\frac{-\pi ik}{5}} - 1}{k} + \sum_{k=1}^{K}\frac{e^{\frac{-\pi ik}{5}} - 1}{k} \right) $$ converges to $-i 4\pi/5$. Add the limiting value $-i\pi/5$ for the $k=0$ term, and get $-\pi$ for your result.