Problem A. 824. (April 2022)

A. 824. An infinite set \(\displaystyle S\) of positive numbers is called thick, if in every interval of the form \(\displaystyle \big[1/(n+1), 1/n\big]\) (where \(\displaystyle n\) is an arbitrary positive integer) there is a number which is the difference of two elements from \(\displaystyle S\). Does there exist a thick set such that the sum of its elements is finite?

Proposed by Gábor Szűcs, Szikszó

(7 pont)

Deadline expired on May 10, 2022.

Solution. Yes, there exists such a thick of set, a suitable construction is for example: let the elements of our set be

• \(\displaystyle 1/n^2\) for each \(\displaystyle n\);
• \(\displaystyle 1/n^2 - 1/m\) for each \(\displaystyle n\) and for each \(\displaystyle n^2<m<(n+1)^2\).

The resulting set is thick: the difference of elements of types 1 and 2 is the reciprocal of all non-squares, and at least one of \(\displaystyle n\) and \(\displaystyle n+1\) is a non-square. On the other hand, the sum of the elements is indeed finite, and to see this it suffices to estimate the sum of elements of type 2, since we know that the reciprocal of the square numbers is finite. For a given \(\displaystyle n\), such members can be estimated from above by \(\displaystyle 1/n^2 - 1/(n+1)^2\), and there are \(\displaystyle 2n\) such members. Their sum is therefore at most \(\displaystyle 2n(2n+1)/(n(n+1))^2<4/n(n+1)<4/n^2\). Thus the sum of all the elements of the set is less than five times the reciprocal of the squares, which is finite. We are done.

Statistics:

3 students sent a solution.
7 points:	Varga Boldizsár.
0 point:	2 students.

Problems in Mathematics of KöMaL, April 2022