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It is well known that the sum $1+2+3+4+\ldots$, which tends to infinity in the regular sense, can be assigned the value $-\frac{1}{12}$ by different means, e.g., zeta regularization or different summation methods.

This question asked if there was a base in which the sum would make sense, though no positive answer was given.

What I was wondering: The notion of convergence depends on the metric we use. Particularly, we would conventionally say that the series must diverge since the terms do not get closer to $0$. But is there a metric where the sequence $1, 2, 3, 4, \ldots$ tends to $0$ such that the series $1+2+3+4+\ldots$ converges, and the limit would indeed be $-\frac{1}{12}$, or possibly some other value? I would think p-adic numbers are a pretty good candidate, but I have very limited knowledge in the field.

Community
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Markus Shepherd
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Here is an example (though it's a bit silly). Let $f : \mathbb{R} \to \mathbb{R}$ be some bijection such that $f(n) = \frac{1}{n}$ and $f(\frac{-1}{12}) = 0$, and define a metric on $\mathbb{R}$ by $d(x,y) = |f(x) - f(y)|$. You can easily check that this satisfies all the axioms of a metric. And $d(1+2+\cdots+n, -1/12) = 1/(1+2+\cdots+n) \to 0$, so for this metric the series $\sum_k k$ converges to $-1/12$.

But as you can see the metric doesn't have a lot to do with the usual metric on $\mathbb{R}$, or with the usual addition/multiplication.

Michael Hardy
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Najib Idrissi
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    Thanks for that! I guess $f(x)$ really only needs to be an injection (not necessarily a bijection), and only for $d(x,y)=0$ iff $x=y$, right? Anyway, this would work for any sequence and any value as the limit. Is there any more natural metric that's specific to this and would offer some insight? – Markus Shepherd Dec 10 '14 at 15:21
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    Is there a non-trivial example, like how $1+2+4+8+\dots$ converges to $-1$ in the 2-adics? I'm tempted to open a new question. – user76284 Dec 04 '18 at 01:25
  • Moreover, how can $f$ be a surjection? What's the pre-image of $-12$? – user76284 Dec 04 '18 at 01:31