TITLE: Erdos Conjecture on arithmetic progressions
QUESTION [26 upvotes]: Introduction:
Let A be a subset of the naturals such that $\sum_{n\in A}\frac{1}{n}=\infty$. The Erdos Conjecture states that A must have arithmetic progressions of arbitrary length. 
Question:
I was wondering how one might go about categorizing or generating the divergent series of the form in the introduction above. I'm interested in some particular techniques and I list some examples below:
If we let $S$ be the set of such divergent series: $S=\left[ A: \sum_{n\in A}\frac{1}{n}=\infty, \ A\in\mathbb{N} \right]$, what kind of operations are there that would make S a group, or at the very least a semigroup? I'm rather vague on what the operatons should be for a reason, because although I presume trivial operations exist, their usefulness in understanding the members of $S$ would be questionable. 
Alternately, can one look at these divergent sums through the technique of Ramanujan summation (think: $1+2+3+\ldots =^R -\frac{1}{12}$, $R$ emphasizing Ramanujan summation)? The generalizations of Ramanujan summation (a good reference  here ) allow one to assign values to some of these series and give some measure of what kind of divergence is occurring. Moreover, basic series manipulations that hold for convergent series tend to carry over to Ramanujan summation, so can one perhaps look at the set $S$ above as a set of equivalence classes in the sense of two elements being equivalent if they share the same Ramanujan summation constant. 
Thanks in advance for any input!

REPLY [3 votes]: An idea that I came up with is that if the conjecture is true, then there would be a finite upper bound on how big the sum could be if the set had no arithmetic progression of length k, lets call this upper bound f(k). Obviously f(1)=0, as if there are any members then there is an arithmetic progression of that length. f(2)=1 (using the set {1}), as adding more numbers would result in an arithmetic progression of length two or more. I am currently working on f(3)