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 $a_2=2$.

Looking at the sequence, we first see that if $a_n' = \alpha a_n$ and $a_{n+1}' = \alpha a_{n+1}$, then also $a_{n+2}' = \alpha a_{n+2}$, and therefore also for the limit $a := \lim_{n\to\infty}a_n$ we have $a' = \alpha a$. Therefore we can write
$$a = a(a_1, a_2) = a_1 a(1, a_2/a_1) =: a_1 f(a_2/a_1).$$
Now what properties does $f$ have? Well, we can of course start the sequence at every position, therefore we have
$$f(x) = a(1, x) = a(x, \sqrt{x}) = x a(1, x^{-1/2}) = xf(x^{-1/2}).$$
Also we can see that if $a_n' = a_n^\alpha$ and $a_{n+1}' = a_{n+1}^{\alpha}$, then also $a_{n+2}' = a_{n+2}^\alpha$, and thus
$$f(x^\alpha) = f(x)^\alpha.$$
Thus we have
$$f(x) = xf(x^{-1/2}) = x(f(x))^{-1/2}$$
and therefore
$$f(x)^{3/2} = x \implies f(x) = x^{2/3}$$
Thus we have
$$a = a_1 f(a_2/a_1) = a_1 \left(\frac{a_2}{a_1}\right)^{\frac23} = a_1^{1/3} a_2^{2/3}.$$
In particular, with $a_1=1$ and $a_2=2$ we get
$$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$
