<s>[INST] Prove that $x_{n+1}=\frac{2}{9}(x_n^3+3)$ converges Let $x_1=1/2$ and $x_{n+1}=\frac{2}{9}(x_n^3+3)$ for $n\geq 1$. We want to prove that the sequence $(x_n)$ converges to real number $r\in (0,1)$ satisfying the equation $2r^3-9r+6=0$.
First part
For the sequence to converge, it must be bounded. A sequence $X:=(x_n)$ of real numbers is said to be bounded if there exists a real number $M>0$ such that $|x_n| \leq M$ for all $n\in \mathbb{N}$.
First, we show that $(x_n)$ converges. 
This is where I struggle


*

*I tried showing by induction that $(x_n)$ < 1 for all $n$** but it didn't work

*I also tried the ratio test...


Second part
Claim  The real number $r$ satisfies the equation $2r^3-9+6=0$
Proof - Since $x_n$ and $x_{n+1}$ convergence to the same real number $r$, we can substitute $r$ in the equation for $x_{n+1}$ when $n$ gets sufficiently large : 
$$
\begin{aligned}
x_{n+1}&=\frac{2}{9}(x_n^3+3)\\
r&=\frac{2}{9}(r^3+3))\\
r&=\frac{2}{9}r^3+\frac{6}{9}\\
0&=\frac{2}{9}r^3-r+\frac{6}{9}\\
0&=2r^3-9r+6 \quad \text{multiply by 9}
\end{aligned}
$$
 [/INST] Induction works fine.  $x_n<1$ implies that $x_n^3<1$.  Hence $$x_{n+1}<\frac{2}{9}(1+3)=\frac{8}{9}<1$$
In fact the same proof shows $\frac{8}{9}$ as an upper bound.
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