What is the limit of a sequence defined recursively as $x_1=2$, $x_{n+1}=1/(3-x_n)$ with $n \in \mathbb{N}$, and how do I prove it exists?

Question

This is an exercise I've found online.

Find the limit of a sequence defined recursively as $x_1=2$, $x_{n+1}=\dfrac{1}{3-x_n}$ with $n\in \mathbb{N}$. Show that the limit exists before attempting to find it.

So far, I have shown that $\{x_{n}\}$ is bounded below by $0$ and above by $2$ since $\frac{1}{3-x_n}>0$ and $\frac{1}{3-x_n}\le 2$ for all $n$.

I'm stuck here because I'm not sure what to show next, and I don't know precisely how to show the limit exists.

Answer 1

At first, you can show that $x_1>x_2$ easily.

If $x_{n-1}>a_n$ holds (and $0\le x_n \le 2$ for all $n$), we get $$\begin{align} x_{n-1}>x_n & \Longrightarrow & -x_{n-1}<-x_n \\ & \Longrightarrow & 3-x_{n-1}<3-x_n \\ & \Longrightarrow & \frac{1}{3-x_{n-1}}>\frac{1}{3-x_n} \\ & \Longrightarrow & x_n > x_{n+1}\\ \end{align}$$

so $x_n>x_{n+1}$. So this sequence convergent, by monotone convergence theorem.

Answer 2

For limit solve the equation $$l=\frac{1}{3-l} \implies l^2-3l+1=0$$ You get $l=\frac{3+\sqrt 5}{2},\frac{3-\sqrt 5}{2}$. Your limit will be $\frac{3-\sqrt 5}{2}$ since $\frac{3 +\sqrt 5}{2} >2$.

Answer 3

If a limit $L$ exists, it must satisfy $$x_{n+1} = \frac{1}{3-x_n} \Rightarrow L = \frac{1}{3-L} \Rightarrow L^2-3L+1=0$$

So that candidates for the limit are $\frac{3\pm \sqrt{5}}{2}$. Since you've proven the series is bounded above by 2, only one of these is possible.