Proving that the sequence $x_n := \sqrt{x_{n-1} x_{n-2}}$ is convergent

Question

Here's the question:

Let $0<a\le x_1 \le x_2 \le b$. Define $x_n := \sqrt{x_{n-1} x_{n-2}}$ for $n\ge 3$. Prove that $(x_n)$ is convergent.

If we show that $(x_n)$ is Cauchy then it is sufficient to conclude that is convergent. Here's what I've been trying to do:

1. Show that $a\le x_n \le b$ for each $n$.
2. Prove that $|x_{n+1}-x_{n} | < r |x_n -x_{n-1}|$ for $n\ge 2$ and some $r<1$.
3. Then finally show that it it indeed Cauchy.

I showed $(1)$ is true. But I'm stuck on $(2)$. Here's what I'm getting $$ |x_{n+1}-x_{n} | = \left| \frac{\sqrt{x_n}}{\sqrt{x_n} + \sqrt{x_{n-1}} }\right| |x_n - x_{n-1}|$$.

I need an estimate on $\left| \frac{\sqrt{x_n}}{\sqrt{x_n} + \sqrt{x_{n-1}} }\right| < r$. I haven't succeeded so far in doing so. Can i get some hints?

Answer 1

Here are the details of the method suggested by Lord Shark the Unknown: let $y_n= \ln (x_n)$. Then $y_n =\frac {y_{n-1}+y_{n-2}} 2$ This gives $|y_n -y_{n-1}|=\frac 1 2 |y_{n-1} -y_{n-2}|$. Iterating this we get $|y_n -y_{n-1}|=\frac 1 {2^{n-2}} |y_2 -y_1|$. It follows that the series $\sum \{y_n-y_{n-1}\}$ is convergent so $\lim y_n$ exists. Since $x_n =e^{y_{n}}$ we see that $\{x_n\}$ is convergent.

Answer 2

$$\frac{\sqrt{x_n}}{\sqrt{x_n} + \sqrt{x_{n-1}}}=1-\frac{\sqrt{x_{n-1}}}{\sqrt{x_n} + \sqrt{x_{n-1}}}\leq 1- \frac{\sqrt{a}}{2\sqrt{b}}<1$$