Real positive number less than one as a series of rationals Let $0<x\leq 1$. Prove that there exists a unique sequence $(p_n)$ of integers such that $1<p_1\leq p_2 \leq\cdots$ and $$x=\sum_{n=1}^\infty\frac{1}{p_1\dots p_n}.$$
For the existence part I completely lost, can someone give an idea?
For the uniqueness part I'm also stuck but I tried the following. Let $$x=\sum_{n=1}^\infty\frac{1}{q_1\dots q_n}$$ be another representation of $x$ such that, for $n_0\in\mathbb{N}$ such that for $n< n_0$, $q_n=p_n$ but $p_{n_0}<q_{n_0}$. Then  it must be
$$
\sum_{n_0}^\infty \frac{1}{p_1\dots p_{n_0}\dots p_n} = \sum_{n_0}^\infty \frac{1}{p_1\dots p_{n_0-1}q_{n_0}q_{n_0+1}\dots q_n}. \tag{$*$}
$$
With this I want to deduce the contradiction $x<x$. To do this I compute
\begin{equation}
\begin{split}
x &= \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots q_{n_0}} + \sum_{n=n_0+1}^\infty\frac{1}{p_1\dots q_{n_0}\dots q_n}\\
&< \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots p_{n_0}}+ \sum_{n=n_0+1}^\infty\frac{1}{p_1\dots q_{n_0}\dots q_n}\\
&< \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots p_{n_0}} -\frac{1}{p_1\dots q_{n_0}} +\sum_{n=n_0}^\infty\frac{1}{p_1\dots p_{n_0}\dots p_n}\\
&= x + \frac{1}{p_1\dots p_{n_0}} -\frac{1}{p_1\dots q_{n_0}},
\end{split}
\end{equation}
where in the second inequality I used $(*)$. Obviously the last part is a number greater than zero $\alpha$, so $x<x+\alpha$. This is the farthest I can get.
 A: Let $p_1$ be the smallest positive integer for which $1/p_1<x.$ Let $x_1 = x,$ and this will be the first term in a sequence of $x\text{s.}$ Then
$$
\require{cancel}
\xcancel{\begin{align}
\frac 1 {p_1} < {} & x \le \frac 1 {p_1-1} \\
& \text{so} \\
0 < x - \frac 1 {p_1} \le {} & \frac 1 {p_1-1} - \frac 1 {p_1} = \frac 1 {p_1(p_1-1)}.
\end{align}}
$$
The above probably needs further work. See the comments below. 
And so we have
$$
0 < (p_1-1)(p_1x-1) \le 1.
$$
Let $x_2 = \text{this latest number}=(p_1-1)(p_1x-1).$
And now $p_2$ is to $x_2$ as $p_1$ to $x_1$ (and here I'm not talking about a ratio; I just mean $p_2$ is found in the same way).
Just keep going like that. That shows existence.
A: Idea for uniqueness: Suppose $q_1>p_1$ and
$$\tag 1 \frac{1}{p_1}+ \frac{1}{p_1p_2}+\cdots = \frac{1}{q_1}+ \frac{1}{q_1q_2}+\cdots.$$
The right side of $(1)$ is then no more than
$$\frac{1}{q_1}+ \frac{1}{q_1^2}+\frac{1}{q_1^3}+\cdots = \frac{1}{q_1-1} \le \frac{1}{p_1}$$
But the left side of $(1)$ is greater than $\dfrac{1}{p_1},$ contradiction. So $q_1>p_1$ is impossible, and by symmetry so is $p_1>q_1.$ Thus $p_1=q_1.$
Possible lather, rinse, repeat scenario.
