# 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?

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?

• won't it be better to mention where the solution should be in? – Jesse P Francis Apr 17 '15 at 5:44
• @Paul you need to be a little more careful than that. See my answer. – nullUser Apr 17 '15 at 5:48

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?

• From here I could assume the limit exists, therefore for every $\epsilon$ I have that $2-\epsilon < u_n$ which means that $2\leq u_n$, but then $u_{n+1}$ would be negative, then the argument fails, right? – Rono Apr 17 '15 at 5:33
• Where do you mean "here"? You don't need to assume the limit exists (this is what you are trying to prove is false. – nullUser Apr 17 '15 at 5:35
• I'll boil down the argument. 1) Show the sequence $u_n$ is well defined. 2) Show that IF the sequence converged, THEN it must converge to $2$, and finally 3) Show that the sequence does not converge to $2$. – nullUser Apr 17 '15 at 5:37
• I mean here on where I have to prove it doesn't converge to 2. I assume it exists and use the previous argument I described in my previous comment to show it wouldn't be possible. – Rono Apr 17 '15 at 5:37
• No no, steps 2) and 3) are completely independent of each other and should not refer to each other at all. You do not get to assume the sequence converges when you are trying to prove that it doesn't converge to $2$. You must show directly that it does not converge to $2$. – nullUser Apr 17 '15 at 5:39

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.

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.