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