Hyperreals - is there a "boundary" between convergent and divergent series? Hypearreals are equivalence classes of sequences of real numbers. Is there a hypperreal number $ h $ such that for every convergent (real) series $ \sum a_n $ we have that the equivalence  class of the corresponding sequence of the terms of this series - $ (a_n) $ - has $ \overline{ |a_n|} \le h $?
In other words, can we find a sort of "boundary" over the hyperreals between convergent and divergent series - with sequences giving convergent series on one side of that boundary and sequences giving divergent series on the other?
If it's not possible in general, can it be done for absolutely convergent series?
EDIT: As shown by JHance, we can't find a bound for sequences producing divergent series, as some of them can belong to the 0 equivalence class.
So the remaining question is - can we find a hyperreal bound for the sequences giving convergent series?
 A: Unfortunately, I don't think any such boundary is possible. In particular, under the usual construction, any non-vanishing, positive, convergent sequence represents an equivalence class strictly greater than 0. But any ultrafilter on $\mathbb{N}$ contains an infinite set $S$ with infinite complement (in particular it contains either the odds or evens). If we let our sequence vanish on $S$, we can put anything in $\mathbb{N}\backslash S$ and still get a sequence in the equivalence class 0. Obviously we could force divergence. 
Edit: When considering whether there is an upper bound on convergent sequences we get the following problem: we certainly can find upper bounds on the set of convergent sequences, but we can't isolate a greatest lower bound. The former is obvious. The latter exemplifies the failure of the least upper bound property to transfer to hyperreals:
Suppose we have a least upper bound $L$. Obviously $L > 0$ so  $L > L/2 > 0$. Thus there is a convergent series $\sum a_n$ (assume positive) such that $L > \overline{\{a_n\}} > L/2$. Finally we see that $\overline{\{2a_n\}} > L $ but $\sum 2a_n$ converges. I suppose we could still ask whether there is a least upper bound up to multiplication by reals.
