I want Show that x is rational if and only if there exists $n\in \mathbb{N}$ for which $x_n$ is an integer. let $x\in\mathbb{R}-\mathbb{Z}$ and define a sequence by $x=\lfloor x \rfloor +\frac{1}{x_1}$ and $k \in \mathbb{N}$ $x_k=\lfloor x_k \rfloor +\frac{1}{x_{k+1}}$
then $x$ is given by 
it easy to see that if $x_n$ is integer then $x_{n-1}$ is rational and by induction $x_1 $ is rational and by that $x$ is rational
but I couldn't prove the converse that if $x$ is rational then there is $n \in \mathbb{N}$ st $x_n$ is an integer
To elaborate more on my comment to the OP, given $a\in\mathbb{Z}$ and $b\in\mathbb{N}$, theEuclidean algorithm to find the greatest common divisor of $a$ and $b$ yields a finite sequence of quotients and residues such that \begin{array}{lcr} a=k_1b+r_1, & k_1\in\mathbb{Z},& 0<r_1<b_1\\ b=k_2 r_1+ r_2, &k_2\in\mathbb{N}, & 0<r_2<r_1\\ \ldots &\ldots & \ldots\\ r_{n-2}=k_n r_{n-1}+ r_n, &k_n\in\mathbb{N}, & 0<r_n<r_{n-1}\\ r_{n-1}=k_{n+1}r_n, &k_{n+1}\in\mathbb{N}. & \end{array} the residue $r_n=g.c.d(a, b)$.
Consequently, if $x=\frac{a}{b}$ and $g.c.d(a, b)=1$ \begin{align} \frac{a}{b}&=k_1+\frac{r_1}{b}\\ &=k_1+\cfrac{1}{\cfrac{b}{r_1}}=k_1+\cfrac{1}{k_2+\cfrac{r_2}{r_1}}\\ &=k_1+\cfrac{1}{k_2+\cfrac{1}{\cfrac{r_1}{r_2}}}=k_1+\cfrac{1}{k_2+\cfrac{1}{k_3+\cfrac{r_3}{r_2}}}\\ &\quad\ldots \\ &= k_1+\cfrac{1}{k_2+\cfrac{1}{\ddots+\cfrac{1}{\ddots+\cfrac{1}{k_{n-1}+\cfrac{1}{k_n}}}}} \end{align}