Rational numbers as a series of rationals

Question

Any real number $0<x\leq 1$ can be written as

$$ x = \sum_{n=1}^\infty \frac{1}{p_1\dots p_n}, $$

where $p_1\leq p_2\leq\dots$ is a unique sequence of integers $>1$. The number $x$ is a rational if and only if for some $n_0\in\mathbb{N}$, $p_n=p_{n_0}$ for all $n>n_0$. I know that for $\pi_k = p_1\dots p_k$ and $x=a/b$,

$$ 0<a\pi_k - b m_k<\frac{b}{p_{k+1}-1} $$

for some $m_k\in\mathbb{N}$.The only if part is easy, I have already done it. However, I do not know how to proceed with the if part. The construction is given here.

EDIT: Here is my try. Since $x=a/b$, we have

$$ 0<a\pi_k - b m_k<\frac{b}{p_{k+1}-1}. $$

Then $\frac{b}{p_{k+1}-1}\rightarrow 0$ unless for some $n_0\in\mathbb{N}$, $\frac{b}{p_{k+1}-1}$ remains finite. That is $p_n=p_{n_0}$ for all $n>n_0$.

Is this claim correct?

Answer 1

Direction 1: ($\Leftarrow$) Assume that for $n\geq n_0$ we have $p_n=p_{n_0}$. Without loss of generality, we may as well also assume that for $n<n_0$, $p_n<p_{n_0}$. Then

$$x=\sum_{n=1}^\infty \frac{1}{p_1p_2...p_n}=\sum_{n=1}^{n_0-2}\frac{1}{p_1p_2...p_n}+\sum_{n=n_0-1}^{\infty}\frac{1}{p_1p_2...p_n}$$

Now, denote

$$\beta=p_1p_2...p_{n_0-1}$$

Then the sum above becomes

$$=\sum_{n=1}^{n_0-2}\frac{1}{p_1p_2...p_n}+\sum_{n=n_0-1}^{\infty}\frac{1}{\beta p_{n_0}^{n-(n_0-1)}}$$

$$=\sum_{n=1}^{n_0-2}\frac{1}{p_1p_2...p_n}+\frac{1}{\beta}\sum_{n=0}^{\infty}\frac{1}{ p_{n_0}^{n}}=\sum_{n=1}^{n_0-2}\frac{1}{p_1p_2...p_n}+\frac{1}{\beta(1-p_{n_0}^{-1})}$$

which is rational.

Direction 2: ($\Rightarrow$) Assume that

$$x=\sum_{n=1}^\infty \frac{1}{p_1p_2...p_n}$$

is rational. Then $x=\frac{a}{b}$ can be written such that $\gcd(a,b)=1$. Then

$$a=\sum_{n=1}^\infty \frac{b}{p_1p_2...p_n}$$

Now, assume by way of contradiction that $p_n$ is not eventually constant. Since $p_n$ is an integer, increasing, and not constant, we know

$$\lim_{n\to\infty}p_n=\infty$$

Let $N$ be the smallest index such that $p_{N}-1>b$

Then we know

$$a\prod_{n=1}^{N-1}p_n=\prod_{n=1}^{N-1}p_n\sum_{n=1}^{N-1}\frac{b}{p_1p_2...p_n}+\prod_{n=1}^{N-1}p_n\sum_{n=N}^\infty \frac{b}{p_1p_2...p_n}$$

$$\Rightarrow a\prod_{n=1}^{N-1}p_n-\prod_{n=1}^{N-1}p_n\sum_{n=1}^{N-1}\frac{b}{p_1p_2...p_n}=\sum_{n=N}^\infty \frac{b}{p_{N}p_{N+1}...p_n}$$

Now, note that the left hand side is an integer as

$$\prod_{i=1}^n p_i\bigg\vert \prod_{n=1}^{N-1}p_n\text{ for }n=1,2,...,N-1$$

The left side is also positive as

$$a=\sum_{n=1}^{\infty}\frac{b}{p_1p_2...p_n}>\sum_{n=1}^{N-1}\frac{b}{p_1p_2...p_n}$$

This implies

$$\sum_{n=N}^\infty \frac{b}{p_{N}p_{N+1}...p_n}\in\mathbb{N}$$

(where we have excluded $0$ from $\mathbb{N}$). But we also know

$$p_N\leq p_{N+1}\leq p_{N+2}\leq \cdots$$

which implies

$$0<\sum_{n=N}^\infty \frac{b}{p_{N}p_{N+1}...p_n}\leq\sum_{n=N}^\infty \frac{b}{p_{N}^{n-N+1}}=\frac{1}{p_N}\frac{b}{1-p_N^{-1}}=\frac{b}{p_N-1}<1$$

This is a contradiction as we have found and integer between $0$ and $1$. We conclude that at some point, $p_n$ becomes a constant sequence.