# Divergent series and integer numbers

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