$u_{n+2}=\sqrt{u_{n+1}u_{n}}$ find limit of $\{u_n\}$ A sequence $\{u_n\}$ is defined as $u_{n+2}=\sqrt{u_{n+1}u_{n}}$   $\forall n \in \Bbb N$ where $0\lt u_1\lt u_2$ 
Prove that  $\{u_n\}$ converges to  $\sqrt[3]{u_{1}u_2^2}$
$\mathbf{MY TRYING}$ Considering two real positive number $u_{n+1}$ and $u_n$ for any  $n \in \Bbb N$ and applying $AM \geq GM $ (I could show every term of the sequence is positive , using induction procedure) , $$\frac{u_{n+1}+u_n}{2}\geq \sqrt{u_{n+1}u_n} = u_{n+2}$$ or,$$u_{n+1}- u_{n+2} \geq u_{n+2} - u_n$$
So,$u_2-u_3 \geq u_3 - u_1$ i.e. $u_1\leq u_3 \leq u_2$   
again,  $u_3-u_4 \geq u_4 - u_2$ i.e.  either     $u_1\leq u_4 \leq u_3 \leq u_2$   or   $u_1\leq u_3 \leq u_2 \leq u_4$ 
Thus we can not get  a specific order in the terms of the sequence.Then, how to proof the convergence of the sequence. Will Cauchy formula be helpful ? 
 A: If you are showing convergence to some exact value, then the AM-GM inequality is not helpful: unless $u_n = u_{n+1}$, which should never occur, you are giving something away with each application.
Instead, consider transforming $u_{n+2} = \sqrt{u_{n+1} u_n}$ by taking the logarithm: $$\log u_{n+2} = \log \sqrt{u_{n+1} u_n} = \frac12 \log u_{n+1} + \frac12 \log u_n.$$
Now this is a much simpler recurrence for the sequence $v_n = \log u_n$, which you should be able to solve; the solution for $u_n$ follows from there.
A: With a change of variables, this becomes a second order linear equation, which can be solved by the usual methods. Take
$$v_n = \ln u_n.$$
Then the relation becomes
$$e^{v_{n+2}} = \sqrt{e^{v_{n+1}}e^{v_n}} \iff v_{n+2} = \frac{1}{2}v_{n+1} + \frac{1}{2}v_n.$$
The characteristic polynomial becomes
$$2r^2 - r - 1 = 0 \iff r = 1 \text{ or }r = -\frac{1}{2}.$$
The general solution must be of the form
$$v_n = A 1^n + B \left(-\frac{1}{2}\right)^n = A + \frac{B(-1)^n}{2^n}.$$
The limit is $A$ as $n \to \infty$. To find $A$, we can substitute $u_1$ and $u_2$ in to get
\begin{align*}
v_1 &= \ln(u_1) = A - \frac{B}{2} \\
v_2 &= \ln(u_2) = A + \frac{B}{4}.
\end{align*}
Adding twice the second to the first,
$$\ln(u_1 u_2^2) = 3A \implies A = \ln\left(\sqrt[3]{u_1 u_2^2}\right),$$
which is the limit of the $v_n$s. The limit of the $u_n$s is therefore,
$$e^A = \sqrt[3]{u_1 u_2^2}.$$
