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.

$$a_{n+2}≥\frac19$$
 $x^2-\frac56x+\frac19=0$ This equation's roots are $\frac16, \frac23$. 
By $a_1=a_2=0$ converge to$\frac16$.
$$\frac16＞a_{n}≥\frac19$$  
and when $a_{k+1}≥a_{k},a_{k+2}≥a_{k+1},$
$$a_{k+3}-a_{k+2}=a^2_{k+2}+\frac{1}{6}\cdot a_{k+1}+\frac{1}{9}-a_{k}$$
              $$≥a^2_{k}+\frac16a_{k}+\frac19-a_k
               ≥(a_k-\frac5{12})^2-\frac1{16}≥0$$
∴$a_n≥a_{n-1}$
