The sequence $\frac{2}{2-u_n}$ diverges Let $(u_n)$ be a sequence defined with $u_{0}$ a real number such that $u_0 \notin \{0,1,2\}$ and $$u_{n+1} = \frac{2}{2-u_n}$$
Prove that $(u_n)$ diverges.
I try to use the fact that this sequence fluctuates, having negatives values followed by values smaller than 1, then getting values bigger than 1 to get a contradiction using the definition of convergence. The problem is that I can't get any additional information after I find a value bigger than 1, because I can't eliminate the possibility that from that point, the sequence will be bound by 2. Am I missing something here? Is there another route I'm not considering?
 A: if not, let $u_n\to u$ as $n\to \infty$, then you will have 
$$u=\frac{2}{2-u}$$
Does it have a solution?
A: First you should show that $u_n$ is well defined for all $n$ if $u_0 \notin \{0,1,2\}$. Hint: induction. If the sequence converged to $u\neq 2$ then you could take the limit of both sides to get $u = \frac{2}{2-u}$ which has solutions $u=1\pm i$. Since sequences of real numbers can't converge to $1\pm i$, it couldn't possibly be that $u_n$ converges unless $u_n \to 2$. Show that $u_n$ does not converge to $2$. Hint: if $0<|u_n-2| \leq \epsilon$, then $|u_{n+1}| \geq 2/\epsilon$ is really big (and therefore NOT close to $2$), can you formalize this?  
A: Let $\mathbb{R}^{*} = \mathbb{R} \cup \{ \infty \}$
and $f : \mathbb{R}^{*} \to \mathbb{R}^{*}$ be the function $f(u) = \frac{2}{2-u}$, one can check that
$$f^{\circ 4}(u) = f(f(f(f(u)))) = u$$
From this, we see unless $f(u) = u$, the sequence $( u_n )$ defined by
$$u_n = \begin{cases} u, & n = 0\\ f(u_{n-1}), & n > 0\end{cases}
\quad\iff\quad
u_n = f^{\circ n}(u) = \underbrace{f(f(\cdots f(u)\cdots)))}_{\verb/iterated /  n \verb/ times/}$$
will be a non-constant periodic sequence.
It is easy to check
$$f(u) = u\quad\iff\quad u = 1 \pm i \notin \mathbb{R}$$
and the exceptional values ${0,1,2}$ in question corresponds to the periodic 
$4$-orbit $$0 \to 1 \to 2 \to \infty \to 0 \to \cdots$$
which contains $\infty$.
From this, we see for any $u \in \mathbb{R} \setminus \{ 0, 1, 2 \}$, the sequence
$( u_n )$ falls within $\mathbb{R}$, non-constant periodic and hence diverges.
A: Let $f(x) = \frac{2}{2-x}$ and consider differencies $|x - f(x)|$. If $u_n$ converges, it should be a Cauchy sequence. If we manage to prove that $|x - f(x)|$ cannot be as close to zero as we want, then the series cannot converge.
Since we are only interested in how close it can be to zero, sign does not matter and we can consider $g(x) = x - f(x) = x - \frac{2}{2-x}$. Then $g'(x) = 1 - \frac{2}{(2-x)^2}$. The derivative is strictly increasing and is zero at $x = 2 + \sqrt{2}$. Since $g(2+\sqrt{2}) = 2 + 2\sqrt{2}$, we conclude that $g(x)$ has a positive global minimum at $2+\sqrt{2}$, so it cannot be arbitrarily close to zero.
