Convergence of $\sum_{n}\frac{q_n}{n}$, where $(q_n)$ enumerates $\mathbb{Q}\cap[0,1]$? This problem is from a book reviewed Sep. 2020 Notices of AMS.

Rational numbers in $[0,1]$ are countable.  Can they be ordered as $(q_n)_n$ so that $\sum_{n=1}^\infty \frac{q_n}{n}$ converges?

My belief is no, since half the rationals are in the upper half of the interval, the tail of the sum will always have half of its terms $\gt \frac{1}{2n}$ implying divergence.  I am having difficulty in making this rigorous.
 A: Here is a proof of  the conjecture made by @ClementC, with an improvement suggested by @TSF.
Theorem. Given any bounded function $f:\mathbb N\to \mathbb R_+$ admitting a subsequence $\{f(n_k)\}_k$ that converges to zero,
there is an enumeration $\{q_n\}$ of the rationals in $[0,1]$ such that
$$
  \sum_nq_nf(n)<\infty.
  $$
Proof. By hypothesis there are arbitrarily small elements in the range of $f$ so we may
choose an infinite subset $N_1\subseteq \mathbb N$ such that $\sum_{n\in N_1}f(n)$ converges.  By discarding infinitely
many elements from $N_1$, if necessary, we may assume that
$$
  N_2:=\mathbb N\setminus N_1
  $$
is also infinite.
Furthermore  let $Q_1$ be the set of rationals defined by
$$
  Q_1=\{1/2^n: n\in \mathbb N\},
  $$
and let $Q_2$ be the complement of $Q_1$ in $\mathbb Q \cap [0,1]$.
All sets so far defined are countably infinite, so there are bijections
$$
  \alpha :N_1\to Q_2,
  $$
$$
  \beta :N_2\to Q_1.
  $$
The union of $\alpha $ and $\beta $ is therefore a bijection,
$$
  \gamma :\mathbb N\to  \mathbb Q \cap [0,1]
  $$
which is
the enumeration we are looking for, that is, $q_n=\gamma (n)$.
We then have  that
$$\begin{align}
  \sum_{n\in \mathbb N} q_nf(n) &=
  \sum_{n\in N_1} \gamma (n)f(n) +   \sum_{n\in N_2} \gamma (n)f(n) \\&=
  \sum_{n\in N_1} \alpha (n)f(n) +   \sum_{n\in N_2} \beta (n)f(n) \\&\leq
  \sum_{n\in N_1} f(n) +   \|f\|_\infty\sum_{n\in \mathbb N} 1/2^n < \infty.
  \tag*{$\blacksquare$}
\end{align}
  $$
A: Ruy makes a good point in the comments you could possibly formalise it as follows.
Consider any ordering of the rational numbers in $[0,1]$ and remove those elements of the form $\frac{1}{n}$ where $n$ is a positive integer which is not a positive power of $2$.
This leaves the sequence $\{a_n\}_{n=1}^\infty$
Then define $$ q_n =  \left\{
                \begin{array}{ll}
                  a_{\log_2 n}, \text{ if n is a positive power of 2}\\
                  \frac{1}{n}, \text{ otherwise}\\
                \end{array}
              \right.$$
This will imply that $$ \displaystyle \sum_{n=1}^{\infty} \frac{q_n}{n} < \sum_{n=1}^{\infty} \left( \frac{1}{n^2} + \frac{1}{2^n} \right) = \frac{\pi^2}{6} + 1$$
