Best Answer:
The terms can be grouped in pairs, so that we have ∑( (1/2n) - (1/√(2n+1)) . Each term is negative, so we consider the negative the of the entire series, and look at ∑( 1/√(2n+1) - (1/2n) ). But for all n > 2, the term 1/√(2n+1) - (1/2n) is greater than 1/2n, which is a known divergent series.

Edit: anonymous dude, I've heard of this Riemann Rearrangement Theorem, which I think borders on saying that some infinite series have "indeterminate" values, if one is indeed allowed to rearrange the terms in any fashion. However, as a practical matter, this "rearrangement" should be limited to only the first n terms, i.e., no "borrowing" from terms that are from the increasingly higher order terms, otherwise, yes, you can get ambiguous or even nonsense results. As an example, the alternating series ±1/n does clearly reach a limit (namely Log(2)) if added strictly sequentially, but this series can be manipulated to sum up to whatever you want.

In my opinion, I think this is one of the soft spots of mathematics, which is supposedly rigorous, unambiguous, and exactly repeatable. It is actually not. I find this kind of ambiguity interesting, because I often wonder if this is the source of ambiguity or uncertainity in physics. But I speculatively digress.

But in any case, please post any interesting results you may come up with, anonymous dude, I would very much like to know.

NOW I give this Y!A question a thumbs up for interest.

Edit 2: I recall now an amusing example of this "Riemann Rearrangment Theorem". Two farmers are planting seeds on two rows, putting in one seed at a time. With every 5th seed that each farmer plants, two birds eats a seed, one from each row. The difference is that the 1st bird eats the seeds sequentially from the beginning of the row, while the 2nd bird eats every 5th seed. Both farmers plant an infinity of seeds. How many are left? In the first case, none are left, while in the 2nd, there is an infinity of them left! This is in spite of the fact both birds are eating the seeds at the same rate.

Edit 3: In fact, the sum ∑( 1/√(2n+1) - (1/2n) ) is suprisingly very close to √(2n) for large n. Maybe I should post this as a Y!A question, "Prove that √(2n) / ∑( 1/√(2n+1) - (1/2n) ) converges to 1 as n -> infinity"

Edit 4: Zo Maar has just answered my Y!A question and has removed "suprisingly" from of my description of this behavior.

Edit 5: Hey, Yellow Cake, I thought it was a stupid question too, but anonymous dude had made this one really interesting. I think it touches one of the problems still unresolved in mathematics, the ambiguity of certain infinite sums. The core of the problem is the idea of looking at an infinity of terms as a "set", with the presumption that there is an definite and unique value for the sum of them. I think it's a fallacy to speak of a sum of an infinite set without attention to order. As a matter of fact, this sort of reminds me of the matter of "Axiom of Choice", see Wiki link.