(Regarding stack exchange users proposing this as a duplicate: I've gone through answers for these types of questions on this site, but I only found proofs for this and no physical meaning if any. And they also did not talk about why the integral test fails for this sum which worked for me all these years and also makes sense. Thanks for reading this)

So, I've known this identity for a long time now but never thought about it much.

$\sum_{r=1}^{\infty}r = -\frac{1}{12}$.

I've also looked at proofs for this sum. One was through the reimann zeta function.

But, what I was wondering about was its intuition.

  • To begin with I am not really that advanced in maths so I just use Integral methods to test the divergence and convergence of series. This makes perfect sense to me. But when testing this series it is obviously not finite. So, why can this be a finite value? Why does the integral test fail here?

  • The second thing is that I am not comfortable with this series. To demonstrate what I am saying, consider I give you apples. Each second, I give you more apples. But this feels like saying the sum of all apples is an orange.

  • Math is simply logic put in the framework of a language (symbols and stuff). Addition, multiplication, division, Integration, Euclidean geometry, Hyperbolic geometry etc.. are just intuition in a language. But sometimes when we obtain absurd results, why are these considered valid? Is there intuition that I am missing? ( that's what I am looking for). Aren't we supposed to check whats wrong with our theory?

  • Is there any intuition (Physical meaning) behind this result that I am not aware of?

  • $\begingroup$ By the way, it is the Riemann zeta function, not the "reimann zeta function". $\endgroup$ Apr 3 '17 at 14:29
  • $\begingroup$ Asking about PHYSICAL meaning of this is more or less by definition off-topic. If this is not a duplicate, then this is off-topic. $\endgroup$
    – Asaf Karagila
    Apr 3 '17 at 14:54
  • $\begingroup$ So is it better if I post this in the physics site? $\endgroup$
    – Chandrahas
    Apr 3 '17 at 15:59

I know about similar results when you have a relative convergent series. It is a series that converge but the series of it's absolute values doesn't converge. For example: $S=1-1+1-1+1...$. Using intuition here we can get: $S=(1-1)+(1-1)+...=0$ or $S=1+(-1+1)+(-1+1)+...=1$ or even worse $S=1-(1-1+1-1+...)=1-S$ that leads to $S=\frac{1}{2}$. Actually, reordering the terms you can get any value you want. Similar happens with $S=\sum(-1)^n\frac{1}{n}$. The main reason of this, it's because them are not absolute convergent series. But I can't imagine a correct demonstration about your statement. Could you give some reference about this.


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