Are the iterates of $f(x) = \frac{3}{2-x}$ bounded? 
Let $\displaystyle u_0=\frac 32$ and $\displaystyle u_{n+1} = \frac{3}{2-u_n}$
Is $(u_n)$ bounded ?

Since $f(x) = \frac{3}{2-x}$ has no fixed point, $(u_n)$ does not converge.
Observation with a computer suggests that this sequence is not bounded. Indeed, it seems that there are terms that get closer to $2$ as $n$ grows. (e.g $u_{829}\approx 2.001)$
Is there a way to prove this formally ?
I haven't been able to make significant progress with this question, so any help is appreciated.
Progress: with $\alpha = 1 + \frac{i}{2\sqrt 2}$ and $\theta  = \arg (1+i\sqrt 2)$, I've proved that $$u_n = \sqrt 3 \frac{\operatorname{Re}(e^{i(n-1)\theta} \alpha)}{\operatorname{Re}(e^{in\theta} \alpha)}$$
Is that any good ?
 A: No, the iterates of $f$ are not bounded. In fact, they are dense in the real line. A nice way to see this is to view the function $f(z)=3/(2-z)$ as a Mobius transformation on the complex plane. A general technique for understanding the iteration of this type of function is described in Section 1.2 of Iteration of Rational Functions by Alan Beardon.
When viewed in this broader context, $f$ has two fixed points, namely $z=1\pm\sqrt{2}i$. We can simplify $f$ by conjugating it with the function
$$\varphi(z)=\frac{z-1+i \sqrt{2}}{z-1-i \sqrt{2}}.$$
Note that $\varphi$ sends one fixed point of $f$ to zero, the other fixed point of $f$ to $\infty$, and it maps the real line onto a circle containing the point
$$\varphi(0)=\frac{-(1-\sqrt{2}i)}{-(1+\sqrt{2}i)}.$$
By "conjugation", I mean that we translate $f$ to the function $S$ defined by
$$
S(z) \equiv \varphi\circ f \circ\varphi^{-1}(z) = \frac{\sqrt{2}-2 i}{\sqrt{2}+2 i}z.
$$
Now, the dynamics of $S$ are very simple. We have $S^n(z)=\lambda^nz$ where $$\lambda=\frac{\sqrt{2}-2 i}{\sqrt{2}+2 i}.$$
As $\lambda$ is a complex number with absolute value $1$ and irrational argument, the iterates of $S$ from any complex number $z_0$ are dense on a circle with radius $|z_0|$. In particular, the orbit under $S$ of $\varphi(3/2)$ is exactly the image under $\varphi$ of the orbit of $3/2$ under the iteration of $f$. Symbolically:
$$S^n\circ\varphi(3/2) = \varphi\circ f^n(3/2).$$
As $S^n\circ\varphi(3/2)$ is dense in a circle, we must have that $f^n(3/2)$ is dense in the real line.
