Limit of $a_{n+2}=a^2_{n+1}+\frac{1}{6}\cdot a_n+\frac{1}{9}$ Find a limit of sequence:
$$a_{n+2}=a^2_{n+1}+\frac{1}{6}\cdot a_n+\frac{1}{9}$$
$$a_1=0,a_2=0$$
I tried to prove that $a_n$ is bounded and monotonic, but I couldn't prove that $a_n$ is monotonic (by strong induction).
So please give a idea how to show that $a_n$ is monotonic.

$ a_3-a_2\ge 0 $.
Assume, $ a_{k+1}-a_k\ge0 $ for every $ k\le n $
then, $ a_{n+2}-a_{n+1}= a_{n+1}^2-a_n^2+(a_{n}-a_{n-1})/6 $.
 which is greater or equal to 0 since
$ a^2-b^2=(a+b)(a-b) $ and by the assumption.
This settles the monotonic part.
