# 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..

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You should exclude $k=0$ – Norbert Apr 8 '12 at 16:22
Ah you are correct. It's done but it looks ugly, a mod should feel free to clean it up if they know a more elegant way. – esproff Apr 8 '12 at 17:20
That way does look better Norbert, although since it's not absolutely convergent I'm worried that not specifying the order in which we sum the terms would be unacceptable. I think I actually might change them both to how Fabian did it. – esproff Apr 8 '12 at 17:32
Ok, I don't mind – Norbert Apr 8 '12 at 17:38

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.

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Alternately: $\sum_{k=1}^{\infty}\cos(\pi k/5)/k$ does converge, and $\sum_{k=1}^{\infty}(-1)/k$ doesn't. – GEdgar Apr 8 '12 at 22:44

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.

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Ah so it doesn't converge, I'm curious what is the general strategy to show this? – esproff Apr 8 '12 at 17:18
@Thoth: you can check out my answer. – Fabian Apr 8 '12 at 17:21
@Fabian: but your answer is only for absolute convergence, am I missing something? – esproff Apr 8 '12 at 17:25
@Thoth: I will add some notes about the sum from $k=1,\dots,\infty$. – Fabian Apr 8 '12 at 17:36