Convergence of $\frac{a_{n+1}}{a_n}$, where $|a_{n+1}a_{n-1} - a_n^2| = 1 $ Let $(a_n)$ a non-decreasing sequence of positive real numbers such that $\lim a_n = \infty$ and $|a_{n+1}a_{n-1} - a_n^2| = 1 $. Prove that the sequence $(\frac{a_{n+1}}{a_n})$ converges.
My little ´´progress´´:
$|a_{n+1}a_{n-1} - a_n^2| = 1 $.
$\left|\dfrac{a_{n+1}a_{n-1}}{a_n^2} - 1\right| = \dfrac{1}{a_n^2} \to 0$
$\dfrac{a_{n+1}a_{n-1}}{a_n^2} \to 1$
$\dfrac{a_{n+1}}{a_n}\dfrac{a_{n-1}}{a_n} \to 1$
 A: We separate two cases.

Case 1: for every $n \geq 0$, we have $a_{n + 1} - a_n \leq 1$.
In this case, since the sequence $(a_n)_n$ tends to infinity by assumption, we have $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1} - a_n}{a_n} = 0$, and it follows that $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1}}{a_n} = 1$.

Case 2: there exists $m\geq 0$ such that $a_{m + 1} - a_m > 1$.
We then have $a_{m + 2} \geq \frac{a_{m + 1}^2 - 1}{a_m} > \frac{a_{m + 1}^2 - 1}{a_{m + 1} - 1} = a_{m + 1} + 1$. Hence by induction, we have $a_{n + 1} - a_n > 1$ for all $n \geq m$. This implies $a_n > a_m + (n - m)$ for all $n > m$.
Without loss of generality, we may assume that $m = 0$ (otherwise just consider the sequence $(b_k)_k$ with $b_k = a_{m + k}$). We then have $a_n > n$ for all $n \geq 1$.
Now rewrite the equation in the question as $\frac{a_{n + 1}}{a_n} - \frac{a_n}{a_{n - 1}} = \pm \frac 1{a_{n - 1}a_n}$. Thus for any $0 < n < N$, we have: $$\frac{a_{N + 1}}{a_N} - \frac{a_{n + 1}}{a_n} = \pm \frac 1{a_na_{n + 1}}\pm\frac1{a_{n + 1}a_{n + 2}}\pm \dotsc\pm \frac1{a_{N - 1}a_N}.$$ Taking absolute value, we get: $$\left|\frac{a_{N + 1}}{a_N} - \frac{a_{n + 1}}{a_n} \right| \leq \sum_{n < k \leq N}\frac1{a_{k - 1}a_k} <\sum_{n < k \leq N}\frac1{(k - 1)k} < \frac 1 n.$$
This shows that the sequence $\left(\frac{a_{n + 1}}{a_n}\right)_n$ is a Cauchy sequence and hence converges.
