11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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? $\endgroup$ – Did Sep 1 '14 at 8:19
  • 3
    $\begingroup$ 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. $\endgroup$ – Hobbyist Sep 1 '14 at 8:22
  • $\begingroup$ In other words: do your work for you. :-( $\endgroup$ – Did Sep 1 '14 at 8:26
  • $\begingroup$ @Did What were your search terms? I found no duplicates. $\endgroup$ – user1729 Sep 1 '14 at 8:35
  • $\begingroup$ @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. $\endgroup$ – Did Sep 1 '14 at 8:43
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Hobbyist Sep 1 '14 at 8:08
  • 1
    $\begingroup$ 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. $\endgroup$ – user1729 Sep 1 '14 at 8:10
  • $\begingroup$ @Hobbyist I think you can use this result in your proof as long as you define rigorously what the limit of this summation means. $\endgroup$ – Taiben Sep 1 '14 at 8:14
  • $\begingroup$ @user1729 If you look at the first equation in the problem statement, the equation doesn't apply to $\theta = 0$. $\endgroup$ – Taiben Sep 1 '14 at 8:16
  • 1
    $\begingroup$ @Taiben Why is that relevant? That just means that the sum make sense, it doesn't help with divergence...(unless I am missing something?) $\endgroup$ – user1729 Sep 1 '14 at 8:20
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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