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 $a_2=2$.

With $b_n = \log a_n$ (as suggested in a comment) the iteration formula
becomes
$$
  b_{n+2} = \frac 12 (b_n + b_{n+1}) \, .
$$
Computing the first iterates numerically leads to the conjecture
that the sequence converges to $(b_0 + 2b_1)/3\, $.
Therefore we define
$$
  b := \frac{b_0 + 2b_1}3 \, , \quad c := \frac{b_1 - b_0}3 \, .
$$
Then
$$
  b_0 = b - 2c \,, \quad b_1 = b + c \\
  b_2 = b - \frac c2\,, \quad b_3 = b + \frac c4 \\
  b_4 = b - \frac c8\,, \quad b_5 = b + \frac c{16} \\
$$
and generally
$$
b_n = b + \frac{(-1)^{n-1}c}{2^{n-1}}
$$
It follows that
$$
  \lim_{n \to \infty} b_n = b = \frac{b_0 + 2b_1}3
$$
and therefore
$$
  \lim_{n \to \infty} a_n = \sqrt[3]{a_0 a_1^2}
$$
In your case ($a_0 = 1$, $a_1 = 2$) the limit is $2^{2/3} \approx 1.587401$.
