Using the Monotone Convergence Theorem to prove convergence of a recursively defined sequence.

Question

Say $x_n=2+\sqrt{x_{n-1}-2}$ and $x_0\geq2$ for $n\in\mathbb{N}$. Use the Monotone Convergence Theorem to prove that $x_n\to2$ or $x_n\to3$ as $n\to\infty$. Specifically, how do I prove that $x_n$ is decreasing?

Answer 1

Hint: If $x_0 = 3$, then $x_n = 3$ for all $n$. Similar for $x_0 = 2$. So assume $x_0 \neq 2, 3$.

$$x_1 = 2 + \sqrt{x_0 - 2}\Rightarrow x_1 - 2 = \sqrt{x_0- 2}$$

So whenever or not $x_1 <x_0$ or $x_1 > x_0$ depends on $x_0$. Note that

$$\begin{cases} x < \sqrt x & \text{if }0<x<1 \\ x > \sqrt x &\text{if } x>1\end{cases}$$

this implies

$$ \begin{cases} x_0 - 2 < \sqrt{x_0 - 2} & \text{if }2< x_0<3 \\ x_0 - 2 > \sqrt{x_0 - 2} &\text{if } x_0>3 \end{cases}$$

Thus $x_0 < x_1$ if $2<x_0<3$ and $x_0 > x_1$ if $x_0 > 3$. In general, we have

$$x_{n+1} = 2 + \sqrt{x_n -2} \Rightarrow (x_{n+1} -2)^2 = x_n -2.$$ Thus

$$(x_{n+1} - 2)^2 - (x_n -2)^2 = x_n - 2 - (x_{n-1} -2) = x_n - x_{n-1}$$

This inequality will be helpful for you to do the induction that $x_n$ is monotone.