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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.

  1. 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.

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    $\begingroup$ If we are proving things from the definition of limit, we may be forced to (re)prove the fact that an increasing sequence which is bounded above has a limit, as is a decreasing sequence which is bounded below. (The sequence of odd-indexed terms is increasing, the sequence of even-numbered terms is decreasing). One can show that in this case the limits are the same. Lots of work. $\endgroup$ Commented Oct 29, 2014 at 14:13
  • $\begingroup$ Maybe this can help. $$\ln(u_{n+1})=\frac{\ln(u_n)+\ln(u_{n-1})}{2}$$, so if $v_{n+1}=\ln(u_{n+1})$ then $$v_{n+1}=\frac{v_n+v_{n-1}}{2}$$. $\endgroup$
    – pointer
    Commented Oct 29, 2014 at 14:24
  • $\begingroup$ @AndréNicolas: Yes, I have noted that. I have actually tried to show that $\displaystyle\lim_{n\to \infty}\left(u_n u_{n+1}^2\right)$ exists if and only if $\displaystyle\lim_{n\to\infty}u_n$ exists. That, when I have tried using (3.) didn't result in something conclusive. Probably some tricky manipulation may help. $\endgroup$
    – user170039
    Commented Oct 29, 2014 at 14:42

1 Answer 1

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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.

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