Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
    
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. –  Ron Jeremy 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. –  Ron Jeremy Apr 8 '12 at 17:32
    
Ok, I don't mind –  Norbert Apr 8 '12 at 17:38
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
Ah so it doesn't converge, I'm curious what is the general strategy to show this? –  Ron Jeremy 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? –  Ron Jeremy 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.