Prove $a_{n+1} = \frac{4+3 a_n}{3+2 a_n}$ is a Cauchy sequence

Question

How to prove that a sequence $a_n$ as defined $a_{n+1} = \frac{4+3 a_n}{3+2 a_n}$ is a Cauchy sequence?

Answer 1

Suppose that $-\sqrt{2} < a_0 < \sqrt{2}$.

You can show that $-\sqrt{2} < x < \dfrac{4+3x}{3+2x} < \sqrt{2}$ for all $x \in (-\sqrt{2},\sqrt{2})$.

So, $-\sqrt{2} < a_n < a_{n+1} < \sqrt{2}$ for all $n$, and thus $\{a_n\}_{n = 0}^{\infty}$ is increasing and bounded above.

What does this tell you? If $a_0 \not\in (-\sqrt{2},\sqrt{2})$, you can still use a similar method.