TITLE: if $\left | x_{n+1}-\frac{x_{n}^2}{x_{n-1}} \right |\leq 1$, show that $(\frac{x_{n+1}}{x_{n}}) $ convergent
QUESTION [13 upvotes]: Let a real positive number sequence $(x_{n})$ such that $\left | x_{n+1}-\frac{x_{n}^2}{x_{n-1}} \right |\leq 1$ and $\sqrt{x_{1}}\geq \sqrt{x_0+1}$. Show that $(\frac{x_{n+1}}{x_{n}}) $ convergent.
I have tried to define a sequence $(y_{n})$ such that $y_{n}=\frac{x_{n+1}}{x_{n}}$. So the preceding inequality equivalent to :
$\left |x_{n}( \frac{x_{n+1}}{x_{n}}-\frac{x_{n}}{x_{n-1}}) \right |\leq 1$
$\equiv \left |y_{n}-y_{n-1}\right |\leq \frac {1}{x_{n}}$
I want to show that $(y_{n})$ is contractive sequence, and then it will be a cauchy sequence if and only if convergent sequence. But i stuck at this part.
Please give me some clue, thank you.

REPLY [4 votes]: What about the following sequence:
$$1,2,3,\frac{7}{2},\frac{37}{12},\frac{865}{504},1,1,\frac{1}{2},1,1,\frac{1}{2},1,1,\frac{1}{2},...$$
I believe it satisfies the given conditions, but clearly $\frac{x_{n+1}}{x_n}$ doesn't converge. I think the problem statement might be wrong.