Evaluate the limit using only the following results

Question

Let $u_2>u_1>0$ and also let $u_{n+1}=\sqrt{u_n u_{n-1}}$ for all $n \geq 2$. Then prove that the sequence $\{u_n\}$ converges.

For this, use of only the following results is permissible,

1. Archimedean Property.

2. For two sequence $\{x_n\}$ and $\{y_n\}$

$$\displaystyle\lim_{n\to\infty}\left(x_n+y_n\right)=\displaystyle\lim_{n\to\infty}x_n+\displaystyle\lim_{n\to\infty}y_n$$

$$\displaystyle\lim_{n\to\infty}\left(x_n\cdot y_n\right)=\left(\displaystyle\lim_{n\to\infty}x_n\right)\cdot\left(\displaystyle\lim_{n\to\infty}y_n\right)$$

provided they exists.

3. For a sequence $\{z_n\}$, the limit $\displaystyle\lim_{n\to\infty}z_n$ doesn't exist is equivalent to saying that, $$\exists \varepsilon>0\mid \left\lvert z_n-l\right\rvert\geq\varepsilon \ \forall l \in \mathbb{R} \land \forall n\geq n_0 (\in \mathbb {N})$$

I have been able to prove that $\displaystyle\lim_{n\to\infty}\left(u_n u_{n+1}^2\right)=u_1u_2^2$. Now one can conclude from 1 and 2 that the limit must be $\sqrt[3]{u_1u_2^2}$ but that happens only if we can prove that the limits exist. And this is exactly where I am stuck. Using only the three mentioned results I can't prove that. Any help will be appreciated.

Answer 1

This is similar to what you was trying, but if it helps... (apologies for my poor English)

Supposing you are allowed to use induction, then $ u_{n+1}^2=u_n u_{n-1}\implies u_{n+1}^2 u_n=u_{n}^2 u_{n-1}=...=u_2^2 u_1 $

Making $ u_2^2 u_1=a $ you have $ u_{n+1}=\sqrt{a/u_n}=\sqrt[4]{a u_{n-1}} $ and again by induction $ u_{n+1}=a^{s_n} u_i^{c_n} $ $ i\in \{1,2\} $, $ s_n $ is a partial sum of a geometric series, and $ c_n $ a power of $\frac{1}{4}$, all depending on the parity of $ n $.

Now you can use the result number $ (3) $ to prove that each element on the right member of the last equality has a limit (I think this is an easier problem than the initial), and then use result $ (2) $ to conclude.