Take some

*infinite* subset of natural numbers

. Define

. For some sets A,

diverges, like the primes

and for other sets,

converges, like the squares

. One could say the more "dense" a set A is, the more likely it is to diverge. One could also compare the "density" of two subsets of

by looking at the asymptotes of their counting functions, i.e. the primes are "denser" than the square numbers because there are more primes under an arbitrarily large x than there are square numbers. But is there such a thing as a "least dense" divergent series?

Question: Let A be a sequence of natural numbers, such that

diverges. Does there exist such an A where no matter how we partition it into two mutually exclusive subsets (

but

=null) either

or

must converge? Find such a set A or prove none exists.

*Conjecture: None exists. For any divergent series A, you will be able to find two mutually exclusive, divergent subsets and .*