This question is motivated by the answer of Gowers to the question Erdos Conjecture on arithmetic progressions.
Question. (1)-Suppose $A \subset \mathbb{N}$ is such that Lim$_n$ $log(n) \cdot |A \cap \{ 1, \dots, n\}|\over{n}$=1. Does $A$ contain arithmetic progressions of arbitrary large finite length?
(2) What if we replace Lim with Limsup?
The answer to both questions is almost certainly yes, but this has not been proven. It remains an open question even for progressions of length 3 (although the current bounds are pretty close).
For progressions of length greater than 3, the best bounds are due to Gowers, which stated in your terms give: if $$ \limsup \frac{(\log\log N)^{c_k}\lvert A\cap \{1,\ldots,N\}\rvert}{N}>0,$$
where $c_k>0$ is some constant depending only on $k$, then $A$ contains infinitely many arithmetic progressions of length $k$. For progressions of length 3 the $(\log\log N)^{c_k}$ can be replaced by $\log N(\log\log N)^{-4}$. For progressions of length 4 it becomes $(\log N)^{c}$, which is a recent result of Green and Tao. Obtaining even this kind of bound, let alone answering your question, for progressions of length 5 or longer seems quite far out of each at the moment.
Based on the constructions we know, something like the following is probably true: if
$$ \limsup \frac{\exp((\log N)^{c_k})\lvert A\cap \{1,\ldots,N\}\rvert}{N}>0$$
then $A$ contains infinitely many arithmetic progressions of length $k$.
