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,
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.
- 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.
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.