<s>[INST] $\lim _{n \rightarrow \infty }a_{n}=\sqrt{a_{n-2}a_{n-1}}$ , $a_1=1, a_2=2$ I was asked to find the limit of:

$a_{n+2}=\sqrt{a_na_{n+1}}$
$a_1=1, a_2=2$

It seems as if the sequence is constant from n=4 and it's value is $a_n=\sqrt{2\sqrt{2}}           -\forall{n>3}$
I'd just like to double-check I did thing right.
Thanks!
 [/INST] Take logs of both sides and define $b_n = \log{a_n}$.  Then
$$2 b_n - b_{n-1}-b_{n-2} = 0$$
$$b_0=0$$
$$b_1=\log{2}$$
This is a constant coefficient difference equation with solution $b_n=A r^n$,where $r$ satisfies
$$2 r^2-r-1=0$$
with solutions $r_+ = 1$ and $r_-=-1/2$.  Thus
$$b_n = A + B \left (-\frac{1}{2} \right )^n$$
with 
$$A+B=0$$
$$A-\frac{1}{2} B = \log{2}$$
The $$b_n=\frac{2}{3} \log{2} \left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]$$
Then going back to the original sequence:
$$a_n = 2^{\frac{2}{3}\left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]}$$
$$\lim_{n \rightarrow \infty} a_n = 2^{2/3}$$
</s>