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

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.

• First note that $$1-x_n=\frac{x_n}{x_{n+1}}$$ implies $$y_n=\prod_{k=1}^n\frac{x_k}{x_{k+1}}=\frac{x_1}{x_{n+1}}$$ Re $(x_n)$, note that $$\frac1{x_{n+1}}=\frac1{x_n}-1$$ hence $$\frac1{x_{n+1}}=\frac1{x_1}-n$$ Can you finish this? – Did Mar 4 '18 at 15:49
• So $\ y_n= \ x_1 - \ {x_1}{n}$ and because of this, by assuming it has a limit, it gives you that limit is -infinity and therefore the assumption is false? Is this approach correct? – Septimiu Cristian Mar 4 '18 at 15:53
• Actually, $y_n=1-nx_1$, from which the divergence to $+\infty$ should be clear without any proof by contradiction. – Did Mar 4 '18 at 15:56
• Yes, I forgot to simplify that x1. Well, thank you! – Septimiu Cristian Mar 4 '18 at 15:57
• ...and then you can accept your own answer. You won't get any points for being the answerer, but it'll mean that the question is "closed out" and doesn't hang there for all eternity on the unanswered-questions list. – John Hughes Mar 4 '18 at 16:01

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.