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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 :

$$ \begin{aligned} x_{n+1}&=\frac{2}{9}(x_n^3+3)\\ r&=\frac{2}{9}(r^3+3))\\ r&=\frac{2}{9}r^3+\frac{6}{9}\\ 0&=\frac{2}{9}r^3-r+\frac{6}{9}\\ 0&=2r^3-9r+6 \quad \text{multiply by 9} \end{aligned} $$

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  • $\begingroup$ Good question! Thanks for showing some work and (especially) letting us know where you're struggling! +1 $\endgroup$ – Tyler Oct 22 '13 at 15:55
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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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  • $\begingroup$ I confused $x_n$ and $x_{n+1}$ when doing it so that I got $x_{n+1} > 1$. Thanks! $\endgroup$ – Justin D. Oct 22 '13 at 13:40
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To correct the first part, a sequence that is bounded need not converge (consider $(-1)^n$). What you can say is that the sequence converges if it is bounded and monotone.

To show monotonicity, show that some $x_{k+1}>x_k$ (try the first two). Now use the fact that $x_{n+1}>x_n$ for some $n$ as your induction hypothesis and show the sequence is increasing in general. Vadim$123$'s argument gives you a clear upper bound. This guarantees convergence, and you have correctly solved the second part.

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