Given the sequence $(a_n)_{n \ge 1}$ with $a_1=2$ and $a_{n+1} = \frac{n^2-1}{a_n} + 2$ for $n \ge 1$, find the following limits. I have the sequence $(a_n)_{n \ge 1}$, such that:
$$a_1 = 2, \hspace{1.5cm} a_{n+1} = \dfrac{n^2-1}{a_n}+2 \hspace{.25cm}, \forall n \ge 1$$
And I have to find $2$ limits:
$$\lim\limits_{n \to \infty} \dfrac{a_n}{n} \hspace{3.5cm} \lim\limits_{n \to \infty} \dfrac{\sum\limits_{k=1}^{n}a_k^3}{n^4}$$
The second one completly put me in the dark. I don't see any trick that I could use.
For the first one, I tried using Stolz-Cesaro:
$$\lim\limits_{n \to \infty} \dfrac{a_{n+1}-a_n}{n+1-n} = \lim\limits_{n \to \infty} (a_{n+1}-a_n) = \lim\limits_{n \to \infty} \bigg ( \dfrac{n^2-1}{a_n} - a_n \bigg )$$
And I got stuck. I don't think I can find a closed form for $a_n$, so I really don't know what should I do to find these $2$ limits.

An easy proof by induction that $n\le a_n\le n+1$ with inductive step$$k\le a_k\le k+1\implies k+1=\frac{k^2-1}{k+1}+2\le a_{k+1}\le\frac{k^2-1}{k}+2<k+2$$implies the first limit is $1$ and the second is $\frac14$, since $\sum_{k=1}^nk^3=\frac14n^4+o(n^4)$. In particular, $1\le\frac{a_n}{n}\le1+\frac1n$ gets the first limit by the squeeze theorem, while$$\begin{align}\frac14\left(1+\frac1n\right)^2&=\frac{\sum_{k=1}^nk^3}{n^4}\le\frac{\sum_{k=1}^na_k^3}{n^4}\le\frac{\sum_{k=1}^n(k+1)^3}{n^4}\\&\le\frac{\sum_{k=1}^{n+1}k^3}{n^4}=\frac14\left(1+\frac1n\right)^2\left(1+\frac2n\right)^2\end{align}$$gives the second by the squeeze theorem. (Here I've used $\sum_{k=1}^nk^3=\frac13n^2(n+1)^2$.)
