Divergence of sequence $y_n=(1-x_1)(1-\ x_2)\cdots(1-x_n)$ if $x_{n+1}=(x_{n+1}+1)x_n$

Question

Let $\ x_n$ be a sequence of real numbers so that $\ x_1$ is between $-1$ and $0$ and $\ x_{n+1}={(\ x_{n+1}+1)}{\ x_n}$ , $n$ is a natural number. Show that $\ y_n= \ (1-\ x_1)(1-\ x_2)\cdots(1-\ x_n)$ is divergent. And I observed that $\ x_{n+1}=\dfrac{\ x_n}{1-\ x_n}$ but I found no reccurencial form for $\ y_n$ or any clue on how to prove it doesn't converge.

Answer 1

In continuation of what I did, we have that $$1-x_n=\frac{x_n}{x_{n+1}}$$ which implies $$y_n=\prod_{k=1}^n\frac{x_k}{x_{k+1}}=\frac{x_1}{x_{n+1}}$$ Re the sequence $(x_n)$, note that $$\frac1{x_{n+1}}=\frac1{x_n}-1$$ hence $$\frac1{x_{n+1}}=\frac1{x_1}-n$$ From this one gets $$y_n=1-nx_1$$ from which the divergence to $+\infty$ follows.