2
$\begingroup$

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 $1/2$.

$\endgroup$
2
$\begingroup$

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.


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).

$\endgroup$
  • $\begingroup$ You're very welcome, but please don't forget to accept either of the answers you've received, were it only for the fact of marking the question as "Answered" (if you find it was). $\endgroup$ – Salvo Tringali Feb 9 '16 at 18:58
1
$\begingroup$

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 $p_{2k-1}\leqslant n<p_{2k+1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.