find the limit for this $(a_{n})_{n \in \mathbb{N}}$, and $a_{0}>0 \\$ and
$a_{n+1}=a_{n}+\frac{1}{a_{n}^2+a_{n}+1}$
Find the limit of: $\lim_{n \to \infty} \frac{a_{n}^3}{n}$
please help me! :D 
Can i solve this whith Cesaro-Stolz theorem or what else?

$$a_{n+1}^3-a_n^3 = (a_{n+1}-a_n)(a_{n+1}^2+a_{n+1} a_{n}+a_n^2) = \frac{a_{n+1}^2+a_{n+1}a_n+a_{n}^2}{a_n^2+a_n+1}$$
and by letting $\delta_n=a_{n+1}-a_n$
$$\begin{eqnarray*} a_{n+1}^2+a_{n+1}a_n+a_n^2 &=& (a_n+\delta_n)^2+(a_n+\delta_n)a_n+a_n^2\\&=&3a_n^2+3\delta_n a_n+\delta_n^2\end{eqnarray*} $$
such that
$$ a_{n+1}^3-a_n^3 = \frac{3a_n^2+\frac{3a_n}{a_n^2+a_n+1}+\frac{1}{(a_n^2+a_n+1)^2}}{a_n^2+a_n+1}.$$
If you manage to prove that $a_n\to +\infty$ the previous line gives $a_{n+1}^3-a_n^3\to 3$, then Stolz-Cesàro proves $a_n\sim\sqrt[3]{3n}$.
Since $x^2+x+1$ has a negative discriminant we have that $a_n^2+a_n+1$ is strictly positive, so the sequence $\{a_n\}_{n\geq 0}$ is increasing and the limit $\lim_{n\to +\infty} a_n$ is either a positive constant $C$ or $+\infty$. On the other hand there is no positive constant $C$ such that
$$ C = C + \frac{1}{C^2+C+1}.$$
