# Evaluate the limit using only the following results

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. 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 $\sqrt{u_1u_2^2}$ 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.

• 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. Oct 29, 2014 at 14:13
• 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 $$v_{n+1}=\frac{v_n+v_{n-1}}{2}$$. Oct 29, 2014 at 14:24
• @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.
– user170039
Oct 29, 2014 at 14:42

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 $u_{n+1}=\sqrt{a/u_n}=\sqrt{a u_{n-1}}$ 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.