Recursive Continued Fraction - $x_{n+1} = 1 + \frac{1}{1+\frac{1}{x_n}}$

Question

I have a sequence defined as such: $x_{n+1} = 1 + \frac{1}{1+\frac{1}{x_n}}$

When calculating the limit of this recursive sequence, I get $$L = \frac{1 \pm \sqrt{5}}{2}$$ I've deduced that if $x_n > 0$ or $x_n <-1$, it must converge to a positive limit, but I am having trouble with $-1 < x_n < 0$. Is it ever possible for such sequence to converge to $- \phi$?

Answer 1

The possible limits of $(x_n)$ are fixed points of $g(x)=\frac{2x+1}{x+1}$. Since $|g'(-\phi)|>1$, the convergence to $-\phi$ is only possible if $x_{n_0} = -\phi$, for some $n_0$.

Answer 2

If $x_0 \in (-1, \frac{1-\sqrt{5}}{2})$ then the sequence is decreasing to $-\infty$. If $x_0 \in (\frac{1-\sqrt{5}}{2}, 0)$ the sequence is convergent to $\frac{1 + \sqrt{5}}{2}.$

Answer 3

Suppose $x_1>0$. Then $x_n>0$ for all $n\ge2$. Let $$ \alpha=\frac{1-\sqrt{5}}{2},\beta=\frac{1+\sqrt{5}}{2}. $$ Since $$ x_2-x_1=1+\frac{1}{1+\frac{1}{x_1}}-x_1=-\frac{x_1^2-x_1-1}{1+x_1}=-\frac{(x_1-\alpha)(x_1-\beta)}{1+x_1} $$ and $$ x_{n+1}-x_n=\frac{1}{1+\frac{1}{x_n}}-\frac{1}{1+\frac{1}{x_{n-1}}}=\frac{x_n-x_{n-1}}{x_{n-1}x_n\left(\frac{1}{1+\frac{1}{x_n}}\right)\left(\frac{1}{1+\frac{1}{x_{n-1}}}\right)}$$ we have that, if $x_1=\beta$, then $\{x_n\}=\{\beta\}$ is a constant sequence; if $x_1<\beta$, $\{x_n\}$ is increasing; and $x_1>\beta$ is decreasing. Thus $\lim_{n\to\infty}x_n$ is either finite or infinite. Clearly the limit is not infinite. Let $\lim_{n\to\infty}x_n=x$. Then $$ x=1+\frac{1}{1+\frac1x} $$ which gives $x=\beta$. If $x_1<0$, you treat similarly and I omit the detail.