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Sorry if this has been posted before. Can somebody please tell me whether this result is correct, and give explanation as to why or why not? I'm not good at the formal side of maths.

Start here: $$\sum\limits_{k=0}^{\infty}e^{ki\vartheta}=\frac{1}{2}+\frac{i}{2}\cot\frac{\vartheta}{2},~0<\vartheta<2\pi.$$

Then equate the real and imaginary parts, so $$\begin{align*}\sum\limits_{k=1}^{\infty}\cos k\vartheta &=-\frac{1}{2},\\ \sum\limits_{k=1}^{\infty}\sin k\vartheta &=0.\end{align*}$$ For $\varphi=\vartheta+\pi$ for $-\pi<\varphi<\pi$ we could write the cosine equation as $\frac{1}{2}-\cos\varphi+\cos 2\varphi-\cdots=0$ which would mean $$1-1+1-1+\cdots=\frac{1}{2}.$$ I'm not a mathematician - is this valid?

Edit: For context, here is why I want this result. If the cosine formula holds and we can integrate it twice to some angle $0<\varphi<\alpha$ then get this interesting result $$\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{1-\cos k\alpha}{k^2}=\frac{\alpha^2}{4}$$ which for the angle of $\pi$ would imply that $$\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}=\zeta(2)-\sum\limits_{k=1}^{\infty}\frac{1}{(2k)^2}=\frac{3}{4}\zeta(2)$$ and finally we get $\zeta(2)=\frac{\pi^2}{6}.$ It's interesting that such a pretty result comes out of what is essentially crappy maths.

Also has that $$1-\frac{1}{4}+\frac{1}{9}-\cdots=\frac{\pi^2}{12}$$ by the way.

Hobbyist
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  • "Sorry if this has been posted before" Right, but why didn't you care to minimally search for the tens of duplicates already on the site? – Did Sep 01 '14 at 08:19
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    I did and I didn't see it. So I apologised in case I missed something, I need to go and do something else right now and won't have time later, didn't have time to search too heavily. It would be more helpful to post the relevant thread though. – Hobbyist Sep 01 '14 at 08:22
  • In other words: do your work for you. :-( – Did Sep 01 '14 at 08:26
  • @Did What were your search terms? I found no duplicates. – user1729 Sep 01 '14 at 08:35
  • @user1729 The part about $1-1+1-\cdots$ is dealt with there (which has several dups already). The part about $\zeta(2)$ was added later on. – Did Sep 01 '14 at 08:43
  • @Did Sure, yeah, there are plenty of posts which talk about 1-1+1-1+..., but I do not think that pointing to these others posts is helpful. Doing so merely says "your result is wrong", but it is nice to say why it is wrong. It is a different error from the standard 1-1+1-1... error – user1729 Sep 01 '14 at 08:47
  • @user1729 You mean, because of the exponential series in the present post? Hmmm, maybe... Note however that the question is "Can somebody please tell me whether this result [that is, $1-1+1-1+\cdots=\frac{1}{2}$] is correct, and give explanation as to why or why not?". This is precisely what the older question I linked to (and some answers here) are doing, hence I might go as far as maintaining that searching a tiny bit the site would have greatly benefitted the OP. – Did Sep 01 '14 at 08:56
  • @Did Yes, precisely, it is not an argument I have come across before (although I do not go out of my way to expose myself to these kinds of arguments). I interpreted "result" as "argument", but that is, perhaps, because I often feel that "a result" is a proof of what you want to show, not what you are actually showing. But that is just me! So I do agree that searching would have helped, but I also maintain that searching is difficult. – user1729 Sep 01 '14 at 09:04
  • @user1729 It is true that I read "result" as "statement of theorem". If one reads it as "result+approach" things are different. (Note that the trick in this post and some variations on it are something Fourier analysts run into all the time.) – Did Sep 01 '14 at 09:09
  • @user1729 And now the fact that an answer omitting the exponential series context is accepted, really makes this a dup. – Did Sep 01 '14 at 13:30
  • If you integrate again, does that give you another sum? – Akiva Weinberger Sep 01 '14 at 13:38
  • @Did Hmmm...it perhaps means that the OPs motives are a duplicate, although I would still argue that the question is not (also, the answer is interesting, as are all answers of the form "it depends", so it is perhaps the best of a not-too-brilliant bunch (personally, I like detail and they are missing it...)). Incidentally, do you know the answer to my comment to Taiben's post? – user1729 Sep 01 '14 at 13:40
  • @user1729 You mean, if some functions $(u_n)$ defined on $(-1,1)$ are such that $\sum\limits_nu_n(x)$ converges to some finite limit $u(x)$ for every $|x|\lt1$, $x\ne0$, and if $|u(x)|\to\infty$ when $x\to0$, does this imply that $\sum\limits_nu_n(0)$ diverges? Dunno... but such a statement probably needs some qualifications to be valid. – Did Sep 01 '14 at 13:56
  • @Did Why the $x\rightarrow 0$? That seems to be what Taiben was talking about, but I am not seeing why its relevant...Maybe I will write down my question properly and post it. Perhaps... – user1729 Sep 01 '14 at 14:00
  • @user1729 Well, I thought this was your question. If it was not, apologies. – Did Sep 01 '14 at 14:02
  • Hi, kind of sensible argument ;-) Try to do some research on the site. It has been asked several time before. A quick search gave me this http://math.stackexchange.com/questions/648665/1-1-1-cdots-frac12 but you will find more. I suggest you search a bit and read what others have already posted. – Umberto Sep 16 '14 at 20:24

4 Answers4

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If by $\sum$ you mean what is usually meant, then $$ \sum_{k=0}^{\infty} e^{ki\vartheta} $$ diverges, and the first formula and the rest of the proof is invalid.

JiK
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This serie doesn't converge in the usual sense (partial sum converging towards a limit), as you can extract sub-sequences that converge towards 1 or 0. But there are alternative definition of summation, like Cesaro or Abel that will make this converge.

Euler spent a lot of time trying to decide wether or not it would make sense to say that this converges.

Wikipedia articles:

1-2+3

Cesaro

Edit: For the record, this is the Dirac comb. It makes sense to admit the convergence to 1/2 if you're thinking of it as a Fourier transform.

Matt B.
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You discovered a very interesting result. Its validity depends on your definition of the summation.

In the usual sense the series is divergent and doesn't have a sum. So it's invalid.

However, your equation is valid if you define the summation to be the Cesaro summation, in which case the limit of the arithmetic mean of the first partial sums of the series is used.

This type of results is widely used in physics, for example, in string theory.

Taiben
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  • Can you please expand on this? I needed this result (particularly the cosine formula) for a quick 'proof' that $\zeta(2)=\frac{\pi^2}{6}.$ I think I ignored that the sum diverges because of the interesting results that follow, but now I'm quite confused why a 'wrong' method gives me pretty results. (Or don't worry, I can look it up myself) – Hobbyist Sep 01 '14 at 08:08
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    Sure, but is the proof still valid then? Equating the real and imaginary parts seems...dodgy, expecially as the series diverges, and it is not clear to me that Cesaro summation's mend this bit. – user1729 Sep 01 '14 at 08:10
  • @Hobbyist I think you can use this result in your proof as long as you define rigorously what the limit of this summation means. – Taiben Sep 01 '14 at 08:14
  • @user1729 If you look at the first equation in the problem statement, the equation doesn't apply to $\theta = 0$. – Taiben Sep 01 '14 at 08:16
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    @Taiben Why is that relevant? That just means that the sum make sense, it doesn't help with divergence...(unless I am missing something?) – user1729 Sep 01 '14 at 08:20
  • Hey just wanting to know, if you know it, could you let me know the name of this sort of result so I can look it up myself later? I'm interested in seeing more of this but I don't know what the name for this swapping definition of sums is. – Hobbyist Sep 01 '14 at 08:33
  • If you google Cesaro summation, you should be able to find the relevant definitions. – Taiben Sep 01 '14 at 08:35
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The equation: $$\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{1-\cos k\alpha}{k^2}=\frac{\alpha^2}{4}$$ seems to be valid for all $a\in[-\pi,\pi]$, so there has to be something gone right with your "proof" (or at least the idea behind it), even if it seems flawed "as-is."

I suspect that your original series are "true" if you view it with the "Cesàro" or "Abel" definition (for all but finitely many points, I hope), which makes the derivation valid. Can someone verify this?

Akiva Weinberger
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