I want to know the limits and convergence/divergence of ${a_{n+1}} = \frac{(a_n^2 + a_n + 2)}{6}$ I`ve the following iteratively defined sequence:
$${a_{n+1}} = \frac{(a_n^2 + a_n + 2)}{6}$$
$${a_1=0}\;\text{for}\;{n\ge1}$$
I want to know the limits and if it is convergent or divergent, but I don't know how to handle two "$a_n$'s" in the formula.
If I have got one "$a_n$" I would find it out by proving monotony and boundedness. I think that I can use the same technique here but I got stuck. Thanks in advance.
 A: As a comment says above, if you want convergence, you can get rid of one of the $a_n$ terms on the right hand side by the following transformation:
$$
a_{n+1} = \frac{a_n^2+a_n+2}{6} = \frac{(a_n + 0.5)^2 + 1.75}{6} \\
\implies (a_{n+1} +0.5)= \frac{(a_n + 0.5)^2 + 4.75}{6} 
$$
Letting $b_n = a_n + 0.5$ gives $b_{n+1} = \frac{b_n^2+4.75}{6}$. 

Show that $b_n$ is a bounded sequence, by induction. I think a bound of $1$ will do(from above, of course, since $b_n \geq 0$ trivially).

Now, note the following fact:

$b_n$ converges to $L \in \mathbb R \iff $for all subsequences $\{b_{n_k}\} \subset \{b_n\}$, there exists a convergent subsequence $\{b_{n_{k_l}}\} \to L$, of $b_{n_k}$.

Take any subsequence $b_{n_k}$. It's bounded, so by Bolzano-Weierstrass contains a convergent subsequence. But where must that subsequence go?
Clearly, the limit must satisfy the recurrence relation (as an equation now) itself. The equation does have a solution : $L \approx 0.938$(It has another solution, but that is $\approx 5.06$, so a convergent sequence is not going there anytime). So, every convergent subsequence goes to $0.938$.
By the fact, since the above is true of any subsequence, we have that $b_n \to 0.938$ and therefore $a_n \to 0.438$
A: One way to address such problems is to look at the function $f(x)$ such that $a_{n+1}=f(a_n)$. In this case
$$f(x)=\frac{x^2+x+2}{6}$$
Because $f'(x)=\frac{2x+1}{6}>0, \forall x\geq 0$, $f(x)$ is ascending. Given $a_2=f(a_1)=f(0)=\frac{1}{3}>0=a_1$ we have $f(a_2)\geq f(a_1)$ which is $a_3\geq a_2$ and by induction $\color{red}{a_{n+1}\geq a_n}$. So, the sequence is ascending. 
Next, if $0\leq x\leq 1$ then $0< \frac{x^2+x+2}{6}\leq \frac{1}{3}<1$ and again, by induction, we can show $\color{red}{0\leq a_n \leq 1}$. So, the sequence is ascending and bounded, thus converging. As a result, the limit of the sequence is the solution of
$$L=f(L) \iff L=\frac{L^2+L+2}{6} \iff L^2-5L+2=0 \Rightarrow L=\frac{5-\sqrt{17}}{2}$$
The other solution $L=\frac{5+\sqrt{17}}{2}>4$ and can be discarded.
