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There's the well-known claim that $$\sum_{n=1}^{\infty} n = -\frac{1}{12} \tag 1$$ Of course in this form, using the usual interpretation of the infinite sum as limit of finite sums, it's wrong, as the sum on the left hand side diverges. The $-1/12$ is obtained by using the series of the zeta function, $$\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$$ and then observing that formally evaluating it for $s=-1$ gives eq. (1).

Now let's consider another "sum" of a divergent series: $$\sum_{n=0}^{\infty} 2^n = -1 \tag 2$$ Again, you can do an argument like the above: If we take the geometric series $$\frac{1}{1-q} = \sum_{n=0}^{\infty} q^n$$ and insert $q=2$, we get the claimed identity.

However, eq. (2) can also be made rigorous by extending the integers to the $2$-adic numbers, where the left hand side indeed converges to $-1$.

My question therefore is:

Does there exist an extension of the integers that makes eq. (1) rigorous in the sense that in that extension the sum actually converges to the value $-1/12$?

celtschk
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    It's not so much that you're extending the integers to the 2-adics, it's that you are equipping the integers with a metric other than the usual one, so convergence means something other than what it usually does. – Gerry Myerson Jul 27 '16 at 09:14
  • @GerryMyerson: Are there examples of divergent series whose regularized value is different from what one gets evaluating some simpler function (assuming it is known) which agrees with the series only for a narrow set of values? In other words, can this "natural" extension be not the "right" one? – Vincenzo Oliva Jul 27 '16 at 09:30
  • Have a look at this reference. A good general book on p-adic numbers is that of N. Koblitz – Jean Marie Jul 27 '16 at 09:52
  • But we can't get the $-1/{12}$ sum with $p$-adics. The terms of the series do not converge $p$-adically to zero for any $p$. – Oscar Lanzi Jul 27 '16 at 10:13
  • @OscarLanzi: The OP is asking if some suitable extension exists, not if some $p$-adics are such a suitable extension. – Vincenzo Oliva Jul 27 '16 at 10:17
  • I know but the comments turned to $p$-adics which is a dead end for the OP's sum. – Oscar Lanzi Jul 27 '16 at 10:35
  • Could you give a pointer to what you calle a "well-known claim" ? – Jean Marie Jul 27 '16 at 11:27
  • "In this form it's wrong" is incorrect: what you mean is "Interpreted in the usual way, it's wrong". It is true in some other interpretations of what an infinite sum means. –  Jul 27 '16 at 13:24
  • @JeanMarie: http://math.stackexchange.com/q/633285/34930 and links therein. – celtschk Jul 27 '16 at 13:25
  • @Hurkyl: I've clarified this point in the question. – celtschk Jul 27 '16 at 13:28
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    I'm pretty sure you could "cheat" and define a topology on the rational numbers specifically for the purpose of making this (or any other series that has no term appearing infinitely often) infinite sum converge (to anything you want). The challenge would be to do something where the topology has wider interest; e.g. making all suitable Dirichlet series converge. –  Jul 27 '16 at 13:38
  • As long as I know instead of extending the integers it's a lot easier to extend the notion of convergence and limits. – AlienRem Aug 10 '16 at 13:25
  • https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Feb 27 '21 at 16:43

1 Answers1

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The $2$-adic topology makes $\mathbb{Z}$ into a Hausdorff topological group (and in fact a topological ring). This is important: the fact that the topology is compatible with addition is used to prove several basic facts about convergent series, e.g. that changing a finite number of terms does not affect convergence.

Claim: There is no Hausdorff topological group structure on $\mathbb{Z}$ such that the sequence $$ \tag{$\star$} \sum_{n=1}^kn,\;\;\;k=1,2,\ldots $$ is Cauchy.

Proof: Suppose such a topology exists. Choose an open neighborhood $U$ of $0$ such that $1\not\in U$, and a neighborhood $V$ of $0$ such that $V-V\subset U$. Here, $V-V=\{v_1-v_2:v_1,v_2\in V\}$. We assume $(\star)$ is Cauchy, so there is some $N>0$ such that $$ \sum_{n=1}^{k_1}n-\sum_{n=1}^{k_2}n\in V $$ whenever $k_1$, $k_2>N$. Now $$ 1=\left(\sum_{n=1}^{N+3}n-\sum_{n=1}^{N+2}n\right)-\left(\sum_{n=1}^{N+2}n-\sum_{n=1}^{N+1}n\right)\in V-V\subset U, $$ which contradicts $1\not\in U$.

Julian Rosen
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