What is the limit of the $a_{n+1}=\frac{1+a_n+{a_{n-1}}^3}{3}$? $a_{n+1}=\frac{1+a_n+a_{n-1}^3}{3}$  with $n \geq 2$  ( $a_1 =0$,  $a_2 = 1/2$)
By mathematical induction, the $0 \leq a_n \leq 1$ and the $a_n$ is monotonically increasing.  If the limit exists, then $1$ or $\frac{-1 + \sqrt{5}}{2}$ would be value.
What is the exact limit of this recurrence relation between two values?
 A: If the limit exists, then let $L = \lim\limits_{n \to \infty} a_n$. We must have $L = \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n + 1} = \lim\limits_{n \to \infty} \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + (\lim\limits_{n \to \infty} a_n) + (\lim\limits_{n \to \infty} a_{n - 1})^3}{3} = \frac{1 + L + L^3}{3}$.
Then we must have $L^3 - 2L + 1 = 0$. Note that $L^3 - 2L + 1 = (L - 1)(L^2 + L - 1)$. Then the three possible values are $L = 1$ and $L = \frac{-1 \pm \sqrt{5}}{2}$.
Note that we have $a_3 = \frac{1 + 1/2}{3} = 1/2$. Then it's easy to show that for all $n$, we have $a_{n + 1} \geq a_n$ using strong induction, since the base cases of $n = 1, 2$ are satisfied and for larger $n$, we have $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} \geq \frac{1 + a_{n - 1} + a_{n - 2}^3}{3} = a_n$.
So the sequence is increasing. Now, we show that the sequence is bounded. In fact, we see that for all $n$, we have $a_n \leq \frac{-1 + \sqrt{5}}{2}$. Let $w = \frac{-1 + \sqrt{5}}{2}$. This can be shown directly for $n = 1, 2$. To show it for other values of $n$, we use strong induction and note that $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + w + w_{n - 1}^3}{3} = w$.
The sequence is bounded and increasing, so it must have a limit. Since the sequence is increasing and $a_2 > \frac{-1 - \sqrt{5}}{2}$, that can't be the limit. And since $w < 1$ and the sequence is bounded by $w$, $1$ cannot be the limit either. Thus, the only possibility for the limit is $w$.
A: I agree with plop's suggestion. To make calculations simpler (not so much, as plop's method is itself simpler), you can try showing that $a_n < \frac{3}{4}$ by induction. For $n=1, 2$ it is clear. Assuming that it is true up to all integers till $n$, consider:
$$\frac{a_{n+1}}{\frac{3}{4}} \leq \frac{4}{3} \frac{1+\frac{3}{4}+(\frac{3}{4})^3}{3} = \frac{139}{144} < 1. $$
This implies $a_{n+1} < \frac{3}{4}$.
A: 
Blockquote

solve x^3 + x + 1 = 3 x
Solution
x = 1,
x = 1/2 (-1 - √5),
x = 1/2 (√5 - 1).
Plot the sequence a_n with Mathematica software, with initial condition

*

*$a_0 = 1, a_1 = 0$


*$a_0 = 0, a_1 = 1$
Both solution converges to golden ratio r = x = 1/2 (√5 - 1)
