Convergence of a certain series. If $(q_n)$ is an enumeration of rationals in $(-1,1)$ except $0$, is the following series convergent?
$$
\sum_{n=1}^{\infty}\frac{1}{(nq_n)^2}
$$
Are there enumerations for which is convergent and enumerations for which is divergent?
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 A: This is to provide an example wherein $q_n \ge h(n)=(1/8)n^{-1/3},$ a slight adjustment of the idea in roger's answer. The factor $1/8$ was inserted just so one could start out the sequence. If this is done, then $1/(nq_n)^2 \le 64/n^{4/3}$ which converges when summed.
Note that $h(n)$ is decreasing, and $h(8^k)=1/2^{k+1}.$
Begin with an enumeration $r_1,r_2,\cdots$ of the rationals in $(0,1)$. We will describe $q_n$ as a rearrangement of the sequence of $r_j$ by a definite rule. 
Step $0$: Take $q_1,q_2,\cdots,q_{2^3}$ as the first eight terms of the $r$ sequence (in their order from that sequence) which are at least $1/2^3.$ Since $h(1)=1/2^3$ and $h$ is decreasing, each $q_j$ is at least $h(j)$ in this first step.
Step $1$: Take $q_{2^3+1},\cdots, q_{2^6}$ as the first $2^6-2^3$ terms of the $r$ sequence not already chosen on step $0$ which are at least $1/2^4$. Since $h(2^3)=1/2^4$ and $h$ is decreasing, each $q_j$ chosen in step 1 is at least $h(j)$.
Each further step ("step $t$") selects another block of the sequence of $q$'s from an index $2^{3t}+1$ through index $2^{3(t+1)}$, and uses the next terms of the $r$ sequence which are at least $1/2^{t+1}$.
Every $r_j$ in the initial list of rationals must eventually get used, since once it becomes available on step $t$ it remains available on all further steps. Then since $r_j$ has a definite position in the initial sequence $r_1,r_2,\cdots$ it will eventually get chosen on some step $t$. 
So this gives an example of such a sequence which is convergent. It is not directly constructive, unless one views the initial sequence of $r's$ as constructive, and the rearrangement process.
A: Let's assume to simplify and WLOG that you only consider the rationals in $(0,1)$. The series is convergent for instance if you take $q_n \geq n^{-1/3}$ for all $n$. It is divergent is you take $q_{2n} \leq \frac 1 {n} $ for all $n$. 
