Is there a set $A$ of positive integers such that


* $\sum\_{n \in A} \frac{1}{n} = \infty$, and
* there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$
which takes infinitely many values in $A$?


**Added on Feb 16, 2015:** Seva answered this question completely.
He proved even a great deal more -- namely, that there is a partition
of $\mathbb{N}$ into


* a set $A$ of asymptotic density $1$ of *'non-values of non-linear polynomials'*,
which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$
of degree $\geq 2$, and
* a set $B$ of asymptotic density $0$ of *'values of non-linear polynomials'*,
which contains all but finitely many positive values taken by any polynomial
$f \in \mathbb{Z}[x]$ of degree $\geq 2$.



Set of integers having finite intersection with the image of any polynomial of degree $\geq 2$
There are countably many polynomials with integer coefficients of degree at least $2$; write them all in a sequence as $P\_1,P\_2,P\_3,...\ $ For every $k\ge 1$, the set $A\_k$ of all positive integers which are *not* the values of any of $P\_1,...,P\_k$ has a divergent sum of reciprocals: $\sum\_{a\in A\_k} 1/a=\infty$. Consequently, we can find pairwise disjoint, finite subsets $A\_k'\subset A\_k$ such that $\sum\_{a\in A\_k'} 1/a>1$. Now let $A:=\cup\_{k\ge 1} A\_k'$. By the construction, the series $\sum\_{a\in A}1/a$ diverges, and for each $k\ge 1$, the image of $P\_k$ is disjoint with $A\_k'\cup A\_{k+1}'\cup\ldots$; hence, has a finite intersection with $A$.


Indeed, a slight variation of this argument shows that there is a subset $A\subset{\mathbb N}$ **of asymptotic density $1$** such that every non-linear polynomial with integer coefficients represents a finite number of elements of $A$ only. To see this, for every $k\ge 1$ fix an integer $N\_k$, and let $B\_k$ be the
set of all positive integers larger than $N\_k$ that are representable by
$P\_k$. We can choose $N\_k$ large enough to have $|B\_k\cap[1,x]|<2^{-k}
x^{2/3}$ for every positive integer $x$, and then the set $B:=\cup\_{k\ge 1}
B\_k$ will have zero asymptotic density. Consequently, the set $A:={\mathbb
N}\setminus B$ will have asymptotic density $1$, and we will also have
$A\cap{\rm Im} P\_k\subseteq[1,N\_k]$ for each $k\ge 1$.


