<s>[INST] Let the sequence $(a_n)_{n\geq 1}$ be given, such that $a_1=a_2=0$ and $a_{n+1}=\frac{1}{3}(a_n+a_{n-1}^2+b),$ find $\lim_{n\to \infty}a_n.$ Let the sequence $(a_n)_{n\geq 1}$ be given, such that $a_1=a_2=0$ and $a_{n+1}=\frac{1}{3}(a_n+a_{n-1}^2+b),$ where $0\leq b<1$. Show that $(a_n)$ is convergent and find $\lim_{n\to \infty}a_n.$
My attempt (trying to prove that $(a_n)$ is bounded):
$$ a_3= \frac{1}{3}b$$
Suppose $0\leq a_n <1$, then we have to prove that, $0\leq a_{n+1}=\frac{1}{3}(a_n+a_{n-1}^2+b)<1.$  We have $a_n,b\in [0,1)$ but what about $a_{n-1}^2$? I'm stuck here. And I'm not sure if this is the right way to find if this sequence if convergent.
 [/INST] Assume $\lim_{n\to \infty}a_n$ exists and equal $l$. 
$l=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\frac{1}{3}(a_n+a_{n-1}^2+b)=\frac{1}{3}l+l^2+b$ or we write $3l^2-2l+3b=0$,
where $0\leq b<1$.
the discriminant is $\Delta:=4-4\cdot3\cdot 3b=4-36b$ but since $0\le b<1$ then $4\ge 4-36b>-32$, which means $\Delta$ possible to have negative values and in this case the sequence is diverge. 
We restrict the assumptions to have $0<\Delta\le 4$, which happens iff $4-36b>0$ i.e., $0<b<\frac{1}{9}$. 
By the general law we have $l_{1,2}=\frac{2\pm \sqrt{\Delta}}{6}$.
Let us estimate $\frac{1}{3}<l_{1,2}=\frac{2\pm \sqrt{\Delta}}{6} \le \frac{2\pm 2}{6}=
\left\{ \begin{array}{l} \frac{2}{3},  \\ 
\\
 0,{\rm{which \,is \, impossible,\,since\, the \, lower\, bound\, is \frac{1}{3}}\,}=l_1,\,\rm{(say)} \\ 
 \end{array} \right.$.
It remains to have a unique value $\frac{1}{3}<l_{2}\le \frac{2}{3}$ depends of the choice of $0<b<\frac{1}{9}$.
so that $a_n \to l_2$, for a specific $\frac{1}{3}<l_{2}\le \frac{2}{3}$.
</s>