Divergence of $\sum_{n=1}^{\infty}\prod_{k=1}^n q_k$ for some enumeration $(q_n)_{n}$ of $\mathbb{Q}\cap (0,1)$ 
Given an enumeration $(q_n)_{n}$ of $\mathbb{Q}\cap (0,1)$, let us consider the series
$$\sum_{n=1}^{\infty}\prod_{k=1}^n q_k.$$

*

*Find an enumeration such that the series is convergent.

*Find an enumeration such that the series is divergent.



*

*is rather easy: take any enumeration $(a_n)_{n}$ of $\mathbb{Q}\cap (0,1/2]$ and any enumeration $(b_n)_{n}$ of $\mathbb{Q}\cap (1/2,1)$. By letting $(q_n)_n=a_1,b_1,a_2,b_2,\dots$, it follows that
$$\sum_{n=1}^{\infty}\prod_{k=1}^n q_k< 2\sum_{n=1}^{\infty}\frac{1}{2^n}=2.$$


*seems to be more challenging and I did not solve it so far. Any hints?
I read about this problem a few years ago, but I can't remember the source. If someone find it please let me know!

Bonus question 1'. Is there any enumeration $(q_n)_{n}$ of $\mathbb{Q}^+$ such that the series is convergent?

 A: Fix an enumeration of $X = \mathbb{Q}\cap(0,1)$. Using this, we construct an ordered list $\mathtt{L}$ by the following algorithm:

*

*set $\mathtt{L}_0 = () $.


*For each $k$ in the list $\mathbb{N}_{1} = (1,2,3,\dots)$,

*

*Pick the first element $q$ of the set $X \setminus \mathtt{L}_{k-1}$ under the prescribed enumeration.


*Pick $N_k \in \mathbb{N}_{1}$ such that $(N_k - \left|\mathtt{L}_{k-1}\right|) \left( q \prod_{r \in \mathtt{L}_{k-1}} r \right) > 1 $.


*Write $m = N_k - \left|\mathtt{L}_{k-1}\right| - 1$ and pick $s_1, \dots, s_m \in X\setminus(\mathtt{L}_{k-1}\cup\{q\})$ such that
$$ (N_k - \left|\mathtt{L}_{k-1}\right|) \left( s_1 \dots s_m q \prod_{r \in \mathtt{L}_{k-1}} r \right) > 1 \tag{*} $$


*Set $\mathtt{L}_k$ as the concatenation of $\mathtt{L}_{k-1}$ and $(s_1, \dots, s_m, q)$. (Note: We have $\left|\mathtt{L}_{k}\right| = N_k$.)
This algorithm clearly exhausts all the elements of $X$. Moreover, it defines a list $\mathtt{L}$ as the 'limit' of $\mathtt{L}_k$'s such that $\mathtt{L}$ enumerates $X$. Finally, if we write $\mathtt{L}=(q_n)_{n\geq 1}$, then by $\text{(*)}$,
$$ \sum_{n=1}^{\infty} \prod_{j=1}^{n} q_j
\geq \sum_{k=1}^{\infty} (\left|\mathtt{L}_{k}\right| - \left|\mathtt{L}_{k-1}\right|) \prod_{j=1}^{\left|\mathtt{L}_k\right|} q_j > \sum_{k=1}^{\infty}1 = \infty. $$

Answer to Bonus Question. Fix an enumeration of $X=\mathbb{Q}\cap(0,\infty)$. Then run the following algorithm:

*

*Set $\mathtt{L}_0 = ()$.


*For each $n$ in $(1,2,3,\dots)$,

*

*Pick the first element $b_n$ of $X\setminus\mathtt{L}_{n-1}$ under the prescribed order.


*Pick an element $a_n$ of $X\setminus(\mathtt{L}_{n-1}\cup\{b_n\})$ such that $a_n \leq \frac{1}{2}$ and $a_n b_n \leq \frac{1}{4}$.


*Set $\mathtt{L}_n$ as the concatenation of $\mathtt{L}_{n-1}$ and $(a_n, b_n)$.
Now let $\mathtt{L}$ be the 'limit' of $\mathtt{L}_n$ as $n\to\infty$. (Regarding $\mathtt{L}_n$'s as partial functions from $\mathbb{N}_1$ to $X$, this amounts to taking the union of all $\mathtt{L}_n$'s.) By the construction, it is clear that $\mathtt{L}$ is an enumeration of $X$. Moreover, if we write $\mathtt{L}_n=(q_n)_{n\geq1}$, then the construction tells that
$$ \prod_{k=1}^{n} q_k \leq 2^{-n} \qquad \text{for all} \quad n \geq 1, $$
and so, the series $ \sum_{n=1}^{\infty} \prod_{k=1}^{n} q_k $ converges.
A: To solve Q.2:
Lemma 1: For $x\in[0,\frac{1}{2}]$, $\ln(1-x)\geq-2x$.
Proof of lemma 1: Define $f:[0,\frac{1}{2}]\rightarrow\mathbb{R}$
by $f(x)=\ln(1-x)+2x$. For $x\in(0,\frac{1}{2}),$ we have
\begin{eqnarray*}
f'(x) & = & \frac{-1}{1-x}+2\\
 & = & \frac{1-2x}{1-x}\\
 & > & 0.
\end{eqnarray*}
Therefore, $f$ is strictly increasing. Hence, for any $x\in[0,\frac{1}{2}]$,
$f(x)\geq f(0)=0$. That is, $\ln(1-x)\geq-2x$.

Let $r_{n}=1-\frac{1}{2^{n}}$, for $n=1,2,\ldots$. Then,
\begin{eqnarray*}
 &  & \ln(\prod_{k=1}^{n}r_{k})\\
 & \geq & \sum_{k=1}^{n}(-2)\frac{1}{2^{k}}\\
 & \geq & -2.
\end{eqnarray*}
It follows that $\prod_{k=1}^{\infty}r_{k}\geq e^{-2}:=\xi>0.$ Fix
an arbitrary enumeration $\{a_{n}\mid n\in\mathbb{N}\}$ for $[\mathbb{Q}\cap(0,1)]\setminus\{r_{n}\mid n\in\mathbb{N}\}$.
Let $b_{1}=a_{1}$. Choose $n_{1}>1$ such that $n_{1}a_{1}\xi>1$.
Define $b_{2}=r_{1}$, $b_{3}=r_{2},\ldots,b_{n_{1}}=r_{n_{1}-1}$.
Observe that $\prod_{i=1}^{j}b_{i}\geq a_{1}\xi$ for $j=1,2,\ldots,n_{1}$,
so $\sum_{j=1}^{n_{1}}\prod_{i=1}^{j}b_{j}\geq n_{1}a_{1}\xi>1$.
Choose $n_{2}>n_{1}+1$ such that $n_{2}a_{1}a_{2}\xi>2$. Define
$b_{n_{1}+1}=a_{2}$, $b_{n_{1}+2}=r_{n_{1}}$, ... , $b_{n_{2}}=r_{n_{2}-2}$.
Observe that for $\prod_{i=1}^{j}b_{j}\geq a_{1}a_{2}\xi$ for all
$j=1,\ldots, n_{2}$. Therefore, $\sum_{j=1}^{n_{2}}\prod_{i=1}^{j}b_{j}\geq n_{2}a_{1}a_{2}>2$.
Choose $n_{3}>n_{2}+1$ such that $n_{3}a_{1}a_{2}a_{3}\xi>3$. Define
$b_{n_{2}+1}=a_{3}$, $b_{n_{2}+2}=r_{n_{2}-1}$, ..., $b_{n_{3}}=r_{n_{3}-3}$.
Observe that $\prod_{i=1}^{j}b_{j}\geq a_{1}a_{2}a_{3}\xi$ for each
$j=1,\ldots,n_{3}$, so $\sum_{j=1}^{n_{3}}\prod_{i=1}^{j}b_{j}\geq n_{3}a_{1}a_{2}a_{3}\xi>3$.
Continue the process indefinitely, then we obtain a sequence $(b_{n})$
such that the map $n\mapsto b_{n}$ is a bijection from $\mathbb{N}$
onto $\mathbb{Q}\cap(0,1)$. That is, $\{b_{n}\mid n\in\mathbb{N}\}$
is an enumeration of $\mathbb{Q}\cap(0,1)$. Moreover, there exists
a sequence of positive integers $(n_{k})$ with $n_{1}<n_{2}<\ldots$
such that $\sum_{j=1}^{n_{k}}\prod_{i=1}^{j}b_{j}\geq n_{k}a_{1}a_{2}\ldots a_{k}\xi>k$.
Therefore $\sum_{j=1}^{\infty}\prod_{i=1}^{j}b_{j}=\infty$.
