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The set of all integer $\mathbb Z$ is closed respect to the operation of addition, this means if we have two integer numbers $n_1$ and $n_2$, the integer $n_3=n_1+n_2$ is still an integer, so even the sum $$S=\sum_{k=1}^nk^p$$ with $p\in \mathbb N$ must be an integer number. Now we know that using the zeta function regularization we can obtain: $$S_0=\sum_{k=1}^\infty{k^0}=-\frac{1}{2}$$ My question is this: is $\mathbb Z$ not closed respect to the addition when we add an infinite number of integers or the sum $S_0$ is not well defined in the set $\mathbb Z$? Any suggestion to solve this problem is appreciated.

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What exactly is the problem you are trying to solve? – Aryabhata Mar 6 '12 at 7:23
I imagine you do not mean that $\sum k^p$ is an integer when, for example, $p=-1$. Do you mean $\mathbb{N}$, not $\mathbb{Z}$? – André Nicolas Mar 6 '12 at 7:24
As is well known, Euler's proof makes heavy use of non integer numbers so, in a sense, it is only natural that the proposed value for $S_0$ is not an integer either. – Did Mar 6 '12 at 7:54
As you (should) know, $S_0$ has little to do with a sum of infinite [ly many] ones. In these divergent series matters, a line should not be crossed, which is to believe that the proposed regularized value is really the sum of the series in the ordinary sense of convergent series (and you might have crossed that line with your last comment). – Did Mar 6 '12 at 8:25
Do I understand you correctly, that your question is, "which of my(your) two formulations are the correct ones to denote the apparent inconsistency when the notions of set properties are applied to an infinite extension" (or so, much likely one can express this even better) – Gottfried Helms Mar 6 '12 at 11:10
up vote 1 down vote accepted

Addition is a binary operation. It is defined on pairs of integers and, by induction, on finite sets of integers. It is not defined on infinite sets of integers. Now you can define something on infinite sets of integers that is sorta kinda like addition, but it isn't actually addition, and so there's no reason to expect it to do everything addition does. In particular, there's no reason to expect that the result of doing this sorta kinda addition-like thing will result in integers.

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Actually, this is the same case as putting negative number into gamma function. You can't find out this answer from the original definition because it is not defined. But it is 'created' by analytic continuation. – Unem Chan Jul 17 '13 at 4:06

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