0,$ there exists a positive number $p_{\eps}$ such that $e^{(2\log (2p))/(p-1)}<1+\eps$ for each $p>p_{\eps}.$ Take the smallest prime number $p=p(\eps)$ greater than $\max \{p_{\eps}, q_k\},$ and put $K(\eps,q_k)=K(\eps,U):=2p(\eps)^4.$ Then $a_n0$) contains some elements that sum to a square. It is essential that we can only sum distinct elements of $A,$ because, for any nonempty set $A \subset \N,$ there is a sumset $A+A+\dots+A$ which contains a square. In this direction, we can mention the following result of T.~Schoen \cite{sch}: if $A$ is a set of positive integers with asymptotic density $d(A)>2/5$ then the sumset $A+A$ contains a perfect square. For more references on sumset related results see the recent book \cite{tavu} of T.~Tao and V. H.~Vu. A `finite version' of the problem on the `densiest' set whose elements do not sum to a square was recently considered by J.~Cilleruelo \cite{cil}. He showed that there is an absolute positive constant $c$ such that, for any positive integer $N \geq 2,$ there exists a subset $A$ of $\{1,2,\dots,N\}$ with $\geq cN^{1/3}$ elements whose distinct elements do not sum to a perfect square. In fact, by taking the largest prime number $p \leq N^{1/3},$ we see that the set $A:=\{p,p^2+p,2p^2+p,\dots,(p-2)p^2+p\}$ with $p-1$ element is a subset of $\{1,2,\dots,N\}.$ Since any sum of distinct elements of $A$ is divisible by $p,$ but not by $p^2,$ we conclude that no sum of distinct elements of the set $A$ of cardinality $p-1 \geq \frac{1}{2} N^{1/3}$ is a perfect power. Notice that in this type of questions not everything is determined by the density of $B.$ In fact, there are some `large' sets $B$ for which there is a `large' set $A$ whose elements do not sum to an integer lying in $B.$ For example, for the set of all odd positive integers $B=\{1,3,5,7,\dots\}$ whose density $d(B)$ is $1/2,$ the `densiest' set $A$ whose elements do not sum to an odd number is the set of all even positive integers $\{2,4,6,8,\dots\}$ with density $d(A)=1/2.$ On the other hand, taking $B=\{2,4,6,8,\dots\},$ we see that no infinite sequence $A$ as required exists. Moreover, if $B$ is the set of all positive integers divisible by $m,$ where $m \in \N$ is large, then the density $d(B)=1/m$ is small. However, by a simple argument modulo $m,$ it is easy to see that there is no infinite set $A \subset \N$ (and even no set $A$ with $\geq m$ distinct positive integers) with the property that its distinct elements always sum to a number lying outside $B.$ Indeed, if $a_1, \dots, a_m \in \N$ then either at least two of the following $m$ numbers $S_j:=\sum_{i=1}^j a_i,$ where $j=1,\dots,m,$ say, $S_u$ and $S_v$ ($u