Find the limit of given sequence Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$
My try:
$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$.
Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$
$$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$
$\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$
$\Rightarrow x^3-2x+1=0$
and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$ 
So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.
how can i say that the limit is
$\dfrac {-1 + \sqrt {5}}{2}$?
 A: We will prove that all $a_n$ are smaller than ${2 \over 3}=0.6666...$. 
By induction, suppose that $0, 1/2, ... a_{n-1}, a_n < 2/3$.
then $a_{n+1} < {1 + 2/3 + 8/27 \over 3 }= {53 \over 81} < {54 \over 81} = {2\over 3}$
since $a_1 =0<{2 \over 3}$, for all n ,  $0 \leqslant a_n < {2 \over 3}$. 
To prove the convergence we will show that a_n is increasing. Again, by induction,
$a_{n+1} - a_n =  {a_n - a_{n-1} \over 3} + { (a_{n-1} - a_{n-2}) ( a_{n-1}^2 + a_{n-1} a_{n-2} + a_{n-2}^2 ) \over 3} > 0$ where 
$a_4-a_3 = {1 \over 24} > 0 $ and $a_5- a_4 = {1 \over 72} > 0  $ are the two first consecutive positive differences.
We have here a strictly increasing bounded sequence i.e. a convergent one.
So the limit calculus in the question is valid. The limit is $\dfrac {-1 + \sqrt {5}}{2} \approx 0.618$
A: Because $f(x)={1\over 3}(1+x+x^3)$ is a strictly increasing function, compute its derivative, show recursively that that $0<a_n$, remark that if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$.
