Every $x \in (0,1]$ can be represented as $x = \sum_{k=1}^{\infty} 1/{n_k}$, such that $n_{k+1}/n_k\in \{2,3,4\}$

Question

Show that every $x \in (0,1]$ can be represented as $x = \sum_{k=1}^{\infty} 1/{n_k}$, where $(n_k)$ is a sequence of positive integers such that $n_{k+1}/n_k\in \{2,3,4\}$.

Please do NOT reveal the full solution, I'm looking for hints.

So, the first strategy that came to my mind was this: I must start with a representation of $x$ in a base-$n$ expansion and then I show that every term $a_k = x_k/n^k$ where $x \in \{0,1,\cdots,n-1\}$ can be written as a finite sum of $1/{n_k}$'s with the given condition, i.e. $n_{k+1}/n_k \in \{2,3,4\}$.

It doesn't sound like a good strategy because even if I replace every $a_k = x_k/n^k$ with a finite sum of the form $a_k=\sum_{i=1}^{m_k}1/n_{i,k}$ and write the number as $$\sum_{k=1}^{\infty}(\sum_{i=1}^{m_k}1/{n_{i,k}})$$

with the condition that for each fixed $k: {n_{i+1,k}/n_{i,k}} \in {2,3,4}$ there is no guarantee that the same holds when we jump from the last term in the sum of $a_k$ to the first term of the sum of $a_{k+1}$.

Please give me some ideas about how one could tackle a problem like this and what are some good strategies to try.

Thanks in advance

EDIT: A new strategy that just came to my mind:

Start like this:

Take $x \in (0,1]$, then by Archemedean property of the reals, we can find $n_1$ such that $1/n_1 < x$. Now choose $n_2$ such that $n_2/n_1 = \min\{2,3,4\}$ provided that $1/n_1 + 1/n_2 < x$. If such a $n_2$ didn't exist, then take $n_1$ smaller enough that $n_2$ exists.. Keep going on this way and you'll have an increasing sequence of positive terms that is bounded above by $x$.. Therefore the sequence must be convergent.. So it has a supremum and now we must somehow control the supremum to be $x$ and it can be done by decreasing the difference between the sum and $x$. Is this idea good?

Answer 1

Hint: For every $m\ge 1$, define $$S_m:=\big\{\sum_{k=1}^\infty\frac1{n_k}: n_1=m, \frac{n_{k+1}}{n_k}\in\{2,3,4\},~\forall k\ge 1\big\},\quad T_m: \Bbb R\to \Bbb R,\ T_m(x)=mx-1.$$

By definition, for every $m\ge 1$, $T_m(S_m)= S_2\cup S_3\cup S_4:=S$, and for $I:=[\frac1{3},1]$, $$S\subset I, \quad T_2^{-1}(I)\cup T_3^{-1}(I)\cup T_4^{-1}(I) \supset I \tag{1}.$$

Using $(1)$ we can prove that $$S=I\quad\Longrightarrow \quad\bigcup_{m=2}^{\infty}S_m=(0,1]~.\tag{2}$$

Proof of $(2)$(sketch):

Given $x_0\in I$, due to $(1)$, we can define $x_k\in I$ inductively on $k$ by choosing $m_k\in\{2,3,4\}$, such that $x_k:=T_{m_k}(x_{k-1})\in I$. Then it is easy to verify that $x_0=\sum_{k=1}^{\infty}\frac1{\prod_{j=1}^k m_j}\in S.$

Answer 2

Hint: If $x \in [\dfrac{1}{3},1]$ then at least one of $2x-1, 3x-1, 4x-1$ is in $[\dfrac{1}{3},1]$

Secondary hint:

Such a fact could be used to generate an infinite sequence consisting of 2s, 3s and 4s.