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 $1/2$.



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



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> T. Šalát, *On subseries*, Math. Zeitschr. **85** (1964), 209-225.
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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 $\lim\_n \frac{1}{n} \sum\_{k=1}^n \varepsilon\_n = 0$.
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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:



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



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> L. Moser, *On the series $\sum 1/p$*, Amer. Math. Monthly **65** (1958), 104-105,
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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](https://mathoverflow.net/questions/211174/who-was-were-the-first-to-note-that-if-sum-x-in-x-frac1x-infty-the) (that's why I remembered of this story, I guess).


