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Given any increasing positive sequence $\{a_n\}$ diverging to infinity, is it possible to construct a non-negative sequence $\{b_n\}$ so that $\{b_n\}$ is summable but $\{a_n b_n\}$ is not? In other words, can we construct two series with arbitrarily small ratio growth but only one of them diverges?

(Edited) is it possible to find b_n so that {b_n} and {a_n b_n} are decreasing?

KingofHeadbutt
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[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_{n_{k}} >k$ Let $b_n=\frac 1 {ka_n}$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $\sum a_nb_n =\sum \frac 1 k =\infty$ and $\sum b_n <\sum \frac 1 {k^{2}} <\infty $. If you want $b_n$'s to be strictly positive simply take $b_n=\frac 1 {n^{2}}$ when $n$ is not of the type $n_k$.

Kavi Rama Murthy
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  • This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $\frac{1}{ka_n}$ equally among the $b_{n_k},\ldots,b_{n_{k+1}-1}$), but that will be more cluttered as one can see just by looking at nested index I wrote there. – Arthur Jul 18 '18 at 08:05
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    @Arthur see my edited answer. – Kavi Rama Murthy Jul 18 '18 at 08:15
  • Oh, I was imagining problems which weren't there. Cool. – Arthur Jul 18 '18 at 08:21
  • Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible? – KingofHeadbutt Jul 18 '18 at 08:42