Sum of subsequence, over index set of non-zero density, of monotone divergent sum also divergent?

Question

For a subset $S$ of the natural numbers $N$ and $n\in N$ let $|S\cap n|$ be the number of members of $S$ that are less than $n$. Suppose $S$ does not have upper asymptotic density $0$. That is, $$0<\lim_{m\to \infty} \sup_{n>m}\frac {|S\cap n|}{n}.$$ Suppose $(x_n)_{n\in N} $ is a decreasing sequence of positive reals such that $\sum_{n\in N}x_n=\infty.$ Is it necessary true that $\sum_{n\in S}x_n=\infty$?

I have tried to construct a counter-example, and I have also tried to prove it, and gotten nowhere at all.

This is motivated by a Q on MathExchange: If $(a_n)_n$ and $(b_n)_n$ are decreasing positive real sequences such that $\sum_na_n$ and $\sum_nb_n$ are divergent, is it possible that $\sum_n\min (a_n,b_n)$ converges? If the answer to my Q is "yes" then the answer to that Q is "no" because at least one of $\{n:a_n\leq b_n\},\;\{n:b_n\leq a_n\}$ has upper asymptotic density of at least $1/2$.

Answer 1

Summary. The answer to your question is "No". But it is "Yes" under the additional condition that $\liminf_n n x_n > 0$. All of this follows from work of T. Šalát in the 1960s. You find some details below.

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Earlier this morning, I had posted another answer (now deleted). But on my way to the chocolate shop I've realized it answered a different (and much easier) question I must have dreamed of...

In addition, and what is perhaps more interesting, I remembered that Georges Grekos had told me of work by Šalát on the very question in the OP. Here is a reference:

T. Šalát, On subseries, Math. Zeitschr. 85 (1964), 209-225.

There, Šalát proves, among other things, the following:

Theorem. Let $(a_n)_{n \ge 1}$ be a non-increasing sequence of non-negative real numbers such that $a_n \to 0$ as $n \to \infty$ and $\liminf_n n a_n > 0$. If $(\varepsilon_n)_{n \ge 1}$ is a $\{0,1\}$-valued sequence for which $\sum_{n \ge 1} \varepsilon_n a_n < \infty$, then $\lim_n \frac{1}{n} \sum_{k=1}^n \varepsilon_n = 0$.

This is Theorem 1 in Šalát's paper, which also investigates the logical strength of the assumptions made in the previous statement. In particular, Note 2 on p. 211 shows that, at least in general, you can't replace the hypothesis that $\liminf_n n a_n > 0$ in the above theorem with the weaker condition that $\sum_{n \ge 1} a_n = \infty$. For this, Šalát considers the sequence $(a_n)_{n \ge 1}$ defined by letting $a_{n^n + k} := n^{-(n+2)}$ for all $n, k \in \bf N$ such that $0 \le k < (n+1)^n - n^n$.

Incidentally, Šalát mentions that his theorem is actually a generalization of previous results from:

J. Krzyś, A theorem of Olivier and its generalizations, Prace math. 2 (1956), 159-164 (in Polish)

and

L. Moser, On the series $\sum 1/p$, Amer. Math. Monthly 65 (1958), 104-105,

which focus on the case where $a_n = \frac{1}{n}$ for all $n$.

For the record, the "theorem of Olivier" alluded to in the title of Krzyś's paper is the same for which Igor Rivin has provided a reference here (that's why I remembered of this story, I guess).

Answer 2

The answer to MathExchange question is yes, hence no for your question. Denote $p_k=(k!)^2$. Define $a_n=1/p_{2k+2}$ if $p_{2k}\leqslant n<p_{2k+2}$, $b_n=1/p_{2k+1}$ if $p_{2k-1}\leqslant n<p_{2k+1}$.