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?

^What if $a_0 = -\dfrac{17}{12}$, or $a_0 = -\dfrac{99}{70}$, or $a_0 = -\dfrac{577}{408}$, or ...? — JimmyK4542, Jul 07 '14 at 01:28
@JimmyK4542: You get exactly the same limit, with some negative initial values. Oh, wait... Update that to "as long as it's not $-3/2$ or $-\sqrt{2}$ (the negative fixed point)". — Eric Towers, Jul 07 '14 at 01:35
If $a_0 = -\dfrac{17}{12}$, then $a_1 = -\dfrac{3}{2}$, and $a_2$ is undefined. Similarly, if $a_0 = -\dfrac{99}{70}$, then $a_1 = -\dfrac{17}{12}$, $a_2 = -\dfrac{3}{2}$, and $a_3$ is undefined. — JimmyK4542, Jul 07 '14 at 01:36
Thank you all for the quick responds ... a_1 = 1 is given and {a_n} is a sequence of real numbers where all n belongs to Z+ — Malith, Jul 07 '14 at 01:51
Just a suggestion, you might want to edit your question to include those details. — JimmyK4542, Jul 07 '14 at 02:41

score 3 · Answer 1 · answered Jul 07 '14 at 01:40

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.

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