Proving that a sequence has $a_n = a_{n+2}$ for $n$ sufficiently large. The problem states:
a sequence of real numbers $a_0, a_1, \dots $ is defined as follows. $a_0$ is an arbitrary real number and for $n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}$ (where $\{x\} = x - \lfloor x \rfloor$).
Prove that for $a_n = a_{n+2}$ for $n$ large enough.
my attempt:
I notice that this is a decreasing sequence because $$\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} < 1$$
since $\{ a_{n+1} \} \le 1$ (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis).
Moreover this sequence is bounded between $0$ and $a_0$ so by the monotone convergence theorem it must have a limit.
Taking the limit I obtain $$\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor  a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor  a_n \rfloor) $$
And so if I prove that there exists $n$ s.t. $ \lfloor a_n \rfloor < 1$ I am done. Because then the limit could only be $0$ and this implies that there exist infinite $a_n = a_{n+2} = 0$.
But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed?
Edit:  Add original text of the problem.
 A: Assuming that $a_0 \gt 0$, note that for $a_n \gt 1$, $a_{n+1} \lt \lfloor a_n \rfloor$, so $\lfloor a_n+1\rfloor \le \lfloor a_n\rfloor -1$ so the sequence is bounded above by $a_n, a_n-1, a_n-2\dots 1$ and must reach $0$ and be constant there.  
Now assume $a_0 \lt 0$.  We have $a_1=\lfloor a_0 \rfloor \{a_0\} \gt \lfloor a_0 \rfloor $ so the sequence cannot go below $\lfloor a_0 \rfloor $.  If $\{a_n\} \lt 1-\frac 1{\lfloor a_n \rfloor }, \lfloor a_{n+1} \rfloor \gt \lfloor a_n \rfloor$ and we proceed toward zero.  If $\{a_n\} = 1-\frac 1{\lfloor a_n \rfloor }, a_{n+1}= a_n$ and we are at a fixed point.  In that case we will have $a_{n+2}=a_n$.  If  $\{a_n\} \gt 1-\frac 1{\lfloor a_n \rfloor },\{a_{n+1}\} \lt 1-\frac 1{\lfloor a_{n+1} \rfloor }$ and we will proceed toward zero next step.  Once we get to $a_n \gt -1$ we have $a_{n+1}=(-1)(a_n+1)=-a_n-1 \gt -1, a_{n+2}=-a_{n+1}-1=-(-a_n-1)-1=a_n$ and we are done.
A:  All positive $a_0$ iterates to zero in finite time, see the graph of the iterating map. For negative $a_0$ there are fixed points at every $x=-m^2/(m+1)$, $m\geq 1$ (period 1 orbits). Either $a_0<-1$ eventually lands at one of those fixed points (which are unstable for $m\geq 2$), or it maps to the interval $[-1,0[$ after a finite number of iterations. The map is here $f_{-1}(x)=-(x+1)$. $-1$ maps to $0$ which is a fixed point, but every $x\in (-1,0)$ maps to $-(1+x)$ and then to $x$ (period 2 orbit). The exercise is very neat!
