Proof by Induction for proving that a sequence is bounded

Question

Consider the sequence defined recursively by

$x_{n+1}=4-\frac{1}{x_{n}}$ with $x_{1}=3$.

Prove that $\{x_{n}\}$ is a bounded sequence by showing that $3\le x_{n}\le2+\sqrt{3}$, $\forall n\in\mathbb{N}$.

So I know I am supposed to use induction. I computed the first few elements so $x_{1}=3,x_{2}=\frac{11}{3},x_{3}=\frac{41}{11}, x_{4}=\frac{153}{41}$ and so I know the sequence is increasing.

So I started my proof as follows. We will prove by induction that $3\leq x_{n}\leq 2+\sqrt{3}$

$3\leq x_{1}=3 \leq 2+\sqrt{3}$ which starts our induction.

Next assume that $3\leq x_{k}\leq 2+\sqrt{3}$ which implies that $-3\geq -x_{k}\geq -(2+\sqrt{3})$.

I assume I am supposed to invert next then add 4? But I'm not too sure about the intervals and such.

Help is greatly appreciated.

Answer 1

Hint:

If $3\le x_k\le 2+\sqrt3$,

then $\dfrac13\ge\dfrac1{x_k}\ge\dfrac1{2+\sqrt3}={2-\sqrt3}$,

so $4-\dfrac13\le4-\dfrac1{x_k}\le2+\sqrt3$.