Sylvester expansion of a rational number I've been searching the internet for any help on this issue but there seems to never have been any references to the subject of my homework, "Sylvester expansions".
The first part consists in proving that there exists, for any $x\in]0;1]$, a unique series $(q_n)_n$ satisfying:
$$ \begin{cases}{} \forall k \in \mathbb{N}, q_{k+1} - 1 \geq q_k(q_k-1)  \\ x = \sum\limits_{k=0}^\infty \frac{1}{q_k} \end{cases} (*) $$
The algorithm for creating such a series is given:
let $x_0=x$ and
$$\begin{cases}{} q_{k} = \lfloor\frac{1}{x_k}\rfloor+1 \\ x_k = \frac{1}{q_k}+x_{k+1} \end{cases} (**)$$
The second part allows us to prove that the expansion of the inverse of an integer verifies the inequality as an equality in the first line of the condition to being a Sylvester expansion $(*)$.

The final part is the one I'm stuck on. I have the prove the following statement:
$x$ is rational $\iff$ there exists $k$ such that $\frac{1}{x_k} \in \mathbb{N^*}$
The reciprocal is rather straight forward, but I'm struggling to prove the implication, and have tried a host of methods.
I've tried writing $x$ as a fraction, re-using the previous inequality and using the algorithm $(**)$ to prove it, but none seem to work. I've tried with concrete numerical examples and it seems to work within one or two steps ($x_1$ or $x_2$).
Thanks in advance.
 A: I finally found an answer. We by a quick induction that $x_k = x - \sum\limits_{i=0}^{k-1} \frac{1}{q_i}$. Let us set the integer series $(a_k)_k$, $(b_k)_k$, $(c_k)_k$, $(r_k)_k$ such that:
$$\begin{cases}{} x_{k} = \frac{a_k}{b_k} \\ a_k \wedge b_k = 1\\ b_k = c_k a_k+ r_k \\ 0 \leq r_k \lt a_k \end{cases}$$
We then notice that $q_k = c_k +1$ and 
$$x_{k+1} = \frac{a_k}{b_k}-\frac{1}{q_k}=\frac{a_k-r_k}{(c_ka_k+r_k)(c_k+1)}$$
Depending on whether the numerator and denominator are coprime or not, we have $a_{k+1} \leq a_k-r_k$. Therefore, we can go on to say:
$$r_{k+1} < a_k-r_k$$
By a very quick induction we show that:
$$\forall i \in \mathbb{N}, r_{k+i}<a_k-r_k-r_{k+1}-...-r_{k+i-1}$$
Applying this to $(k,i)=(0,a_0)$ we find that:
$$r_{a_0}<a_0-r_0-...-r_{a_0-1}$$
Suppose that the series $(r_k)_{0\leq k<a_0}$ has all its terms different from 0. They're then all greater than 1 (integer series) and we find that:
$$r_{a_0}<a_0-a_0 \leq 0$$
Which is absurd since the remainders are positive. Therefore, there exists a certain $k \in \mathbb{N}$ such that $r_k=0$ and therefore $x_k = \frac{1}{c_k}$.

I'm fairly certain this is one of the easiest proof that can be provided for this problem, let me know if you find simpler.
