Compute $\lim_\limits{n\to\infty}a_n$ where $a_{n+2}=\sqrt{a_n.a_{n+1}}$ I managed to show that the limit exists, but I don't know how to compute it.
EDIT:
There are initial terms: $a_1=1$ and $a_2=2$.

Note that $$  a_{n+2}\sqrt{a_{n+1}}=a_{n+1}\sqrt{a_n} =\cdots =a_2\sqrt{a_1}=2$$
Hence limit is $2^\frac{2}{3}$. 
