anonymous
  • anonymous
How would you find the sum of this series?
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sum_{n=-\infty}^{+\infty}\frac{\sin(n)}{n}\] I thought it would be better to make another question but with a specific series. I realize there may be multiple ways of finding this sum, but I'm mainly curious about a contor integration way, if one is known by anyone.
tkhunny
  • tkhunny
1) Does it converge? 2) Does it have a sum?
anonymous
  • anonymous
Its sum is \(\pi\). I just don't know how I would get to that conclusion. I may be asking too much that there be a contour integration way to show it, but I was hoping someone might know :)

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anonymous
  • anonymous
Hmm, I keep making some mistake and ending up with \(-\dfrac{1}{2}\)...
anonymous
  • anonymous
I would assume you have to treat it like \[1 + 2\sum_{n=1}^{+\infty}\frac{ \sin(n) }{ n }\] I don't know if that changes the process of evaluating it if you have it in that form, though.
anonymous
  • anonymous
Here's what I have so far: Since \(\dfrac{\sin n}{n}\) is even, you can consider a contour integral with \(f(z)=\dfrac{\sin z\cot\pi z}{z}\) with a square contour that surrounds the poles. You have poles at all the integers, which should be easy to deal with. I forget the details, but the integral \(\displaystyle \oint_C f(z)\,\mathrm{d}z\) should disappear, so you have \[0=\sum_{k\in\mathbb{Z}}\text{Res}(f(z),k)=\text{Res}(f(z),0)+\sum_{k\in\mathbb{Z}\setminus\{0\}}\text{Res}(f(z),k)\](I only split the poles up because there's nice pattern with the non-zero poles' resiudes.) Using the expansion is one way to compute the residues, but since the poles are simple, I'll just compute via the limit route: \[\text{Res}(f(z),0)=\lim_{z\to0}z\frac{\sin z\cot \pi z}{z}=\dfrac{1}{\pi}\]For the rest, you have \[\sum_{k\in\mathbb{Z}\setminus\{0\}}\lim_{\substack{z\to k\\k\in\mathbb{Z}}}(z-k)\frac{\sin z\cot \pi z}{z}=\sum_{k\in\mathbb{Z}\setminus\{0\}}\frac{\sin k}{k\pi }=\frac{2}{\pi}\sum_{k=1}^\infty\frac{\sin k}{k}\]where the last equality holds because \(\dfrac{\sin n}{n}\) is even, and so \[\begin{align*}0&=\text{Res}(f(z),0)+\sum_{k\in\mathbb{Z}\setminus\{0\}}\text{Res}(f(z),k)\\[1ex] -\frac{1}{\pi}&=\frac{2}{\pi}\sum_{k=1}^\infty\frac{\sin n}{n}\end{align*}\]I clearly have made some mistake somewhere, but I'm not sure where.
anonymous
  • anonymous
The residues should be fine, I've been checking back and forth with Mathematica.
anonymous
  • anonymous
Well, I know you need an extra \(\pi\) if the integrals are of that form. \(f(z)\pi \cot(\pi z)\) where our f(z) would be \(\frac{\sin(z)}{z}\), which should make that result come out to be just 1.
anonymous
  • anonymous
Hmm, I was under the impression that the inclusion of \(\pi\) in \(f(z)\) isn't necessary, as it should cancel out... At least that's what happened when I reviewed the Basel problem via contour integration from my old lecture notes.
anonymous
  • anonymous
Am I missing a pole somewhere?
anonymous
  • anonymous
Ah. Well, we were only given two different types of contour integrals, both of which used an extra \(\pi\). But I'm not sure if that changes with series of sine and cosine, we weren't given anything too specific on that, just that it was possible.
anonymous
  • anonymous
I initially thought the idea was to use exponential form, but doing that gave me problems with the contour. I was trying to make sure everything vanished, or at least gave a convergent result, but I was unable to do that.
anonymous
  • anonymous
I'm not familiar with the exponential form (though I doubt it changes much at first glance). I remember coming across a pdf on the web that surveyed only three variations of \(f(z)\): \(\cot\pi z\) for even summands, \(\csc\pi z\) for alternating ones, and \(\sec\pi z\) for odd.
anonymous
  • anonymous
Hm. We were given \(\cot(\pi z)\) and \(\csc(\pi z)\). Good to know there's a \(\sec(\pi z)\) one. As for the exponential form idea, I guess it goes along with how you would integrate functions with sine and cosine. Like to integrate \(\frac{\sin(x)}{x}\) you would look at \(\frac{e^{iz}}{z} \), that's why I had that idea.
anonymous
  • anonymous
Ah here's the file I mentioned. Cached link: http://webcache.googleusercontent.com/search?q=cache:I7_GuQ5MlhAJ:www2.mae.ufl.edu/~uhk/SERIESCOMP.pdf+&cd=10&hl=en&ct=clnk&gl=us
anonymous
  • anonymous
So when I tried that idea, I was considering \[\oint \frac{e^{iz}\pi \csc(\pi z)}{z} dz\] And okay, Ill take a look :)
anonymous
  • anonymous
You would think that it'd be possible to find exact sums of things like \[\sum_{n=1}^{\infty}\frac{1}{n^{3}}\] since \(\sec(\pi z)\) handles some odd functions. Oh well. So you used \(\cot(\pi z)\), which is the idea for even functions. I suppose my thought process was that since sine and cosine alternate that \(\csc(\pi z)\) would work. And maybe it does in the general case, but the evenness of the function takes priority?
anonymous
  • anonymous
It seems to me that the choice of \(\cot\pi z\) is a natural one if \(f(n)\) is even, since the residues would be the same for positive/negative non-zero integers, so much so that it should work regardless of what \(f(n)\) is so long as it's even.
anonymous
  • anonymous
Yeah, and after messing around with it, it seems using \(\csc(\pi z)\) gives a \((-1)^{k}\) when computing residues, so it does no good unless your series truly is alternating. Well, going through my notes, there's a small mention that \(f(z)\) is meromorphic without integer poles, which could be what is messing things up.
anonymous
  • anonymous
So maybe what could be done instead is to consider some other sum. \[\sum_{n=-\infty}^{+\infty}\frac{ \sin(n) }{ n+\alpha }\] where \(\alpha \neq n\) There may be a better sum to consider, but its an idea. If we have the sum of this general series, then we just let \(\alpha =0 \) in the final result.
anonymous
  • anonymous
I mean \(\alpha \neq -n \)
anonymous
  • anonymous
Turns out the problem is with the contour (or possibly the integrand). We've been using the square centered at the origin with vertices at \(\left(\pm\left(N+\frac{1}{2}\right),\pm\left(N+\frac{1}{2}\right)\right)\). We have to have\[\left|\oint_\gamma f(z)\,\mathrm{d}z\right|\le M\,\ell(\gamma)\]where \(M\) is some constant and \(\ell(\gamma)\) is the arc length of the contour. Obviously, \(\ell(\gamma)=8N+4\). However, our \(f(z)\) is not bounded in absolute value. For instance, in the case where \(z=x+i y\) with \(x=N+\dfrac{1}{2}\) and \(|y|<\dfrac{1}{2}\) (the right edge of the contour), we have \[\left|\oint_\gamma f(z)\,\mathrm{d}z\right|\le\oint_\gamma|f(z)|\,|\mathrm{d}z|=\frac{\pi(8N+4)\tanh\frac{\pi}{2}}{\sqrt{(N+\frac{1}{2})^2+y^2}}\to8\neq0\]as \(N\to\infty\).
anonymous
  • anonymous
Oops, didn't even realize this had a response to it, my apologies. But yeah, ML Inequality, eh? So it would've never converged on that contour in the first place, got it. I'm glad you came up with that, it's definitely something I want to keep in mind for the future. Don't suppose you came up with an appropriate contour, did ya?
anonymous
  • anonymous
Nothing comes to mind, but I think you can use the same one if you manipulate the integrand or summand. It's the \(\dfrac{1}{z}\) factor that's making this not work, I think. Integration/summation by parts could provide a way around it.

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