Why do sequences $(x_n)$ of real numbers defined by $x_{n+1}=1+\frac 1{x_n}$ nearly always converge to the golden ratio?

Question

I want to prove that the sequence $(x_n)$ of real numbers defined by $x_{n+1}=1+\frac 1{x_n}$ and $x_0\in\Bbb R\setminus \left\{a_n\ \left|\ a_0=0 \text{ and } a_{n+1}=\frac 1{a_n-1} \right.\right\}$ is convergent. It is not monotonic, because $f(x)=1+\frac 1 x$ is a decreasing function, but I have figured out that $x_{n+1}>x_n$ when $x_n \in \left(-\infty,\frac 1 2 -\frac {\sqrt 5} 2\right)\cup\left(0, \frac 1 2 +\frac {\sqrt 5}2\right)$ with $x_{n+1}>0$ when $x_n<-1$, and $x_{n+1}< x_n$ when $x_n \in \left(\frac 1 2 -\frac {\sqrt 5} 2,0\right)\cup \left(\frac 1 2 +\frac {\sqrt 5} 2,\infty \right)$. The mapping $x\mapsto1+\frac 1 x$ is not a strong contraction, so I cannot use Banach's Fixed Point Theorem.

Secondly, how can I prove that $\frac 1 2 +\frac{\sqrt 5} 2$ is the limit of the sequence when $x_0\neq \frac 1 2 -\frac{\sqrt 5}2$?

Answer 1

The map $x\mapsto 1+{1\over x}$ defined on ${\mathbb R}\setminus\{0\}$ can be extended to a Moebius transformation of the Riemann sphere $\bar{\mathbb C}:={\mathbb C}\cup\{\infty\}$: $$T:\quad \bar{\mathbb C}\to \bar{\mathbb C},\qquad z\mapsto {z+1\over z},\quad T(0)=\infty,\quad T(\infty)=1\ .\tag{1}$$ Its fixed points are $\alpha:={1\over2}(1+\sqrt{5})$ and $\beta={1\over2}(1-\sqrt{5})$, obtained by solving the equation $z^2-z-1=0$.

We now introduce a new complex coordinate $w$ on $\bar {\mathbb C}$, related to $z$ via $$w=\phi(z):={z-\alpha\over z-\beta},\qquad{\rm resp.}\qquad z=\phi^{-1}(w):={\alpha-\beta w\over 1-w}\ .$$ The fixed points now are $w=0$ and $w=\infty$. In fact, in terms of the new coordiate $w$ the transformation $T$ appears as $\hat T=\phi\circ T\circ\phi^{-1}$, and computes to $$\hat T:\quad \bar{\mathbb C}\to \bar{\mathbb C},\qquad w\mapsto{\beta\over\alpha}w,\quad \hat T(0)=0,\quad \hat T(\infty)=\infty\ .$$ Since $${\beta\over\alpha}=-{3-\sqrt{5}\over2}\doteq-0.382$$ we can infer that the fixed point $0$ is attracting with basin of attraction all of ${\mathbb C}$, while $\infty$ is repelling. This allows to conclude that in the original setting all initial points $x_0\ne \beta$ lead to $\lim_{n\to\infty} x_n=\alpha$ (assuming the "exception handling" described in $(1)$).

Answer 2

You can proceed as follows. It will suffice to show that there is an index $r$ such that $x_r> 0$.

If $x_0\in A_0=(0,\infty)$, then $r=0$ and we are done.

If $x_0\in B_0=(-\infty, -1)$, then $r=1$ and we are done.

So we may assume without loss that $x_0\in (-1,0)$.

If $x_0\in A_1=\left(-\frac{1}{2},0\right)$, then $x_1\in B_0$, $r=2$ and we are done. So we may assume without loss that $x_0\in\left(-1,-\frac{1}{2}\right)$.

If $x_0\in B_1=\left(-1,-\frac{2}{3}\right)$, then $x_1 \in A_1$, $r=3$ and we are done. So we may assume without loss that $x_0\in\left(-\frac{2}{3},-\frac{1}{2}\right)$.

Continuing this way, we obtain (for $n\geq 1$) the two families $A_n=\left(-u_{n+1},-u_{n}\right), B_n=\left(\frac{-1}{u_{n}+1},\frac{-1}{u_{n+1}+1}\right)$ where $(u_n)$ is defined by $u_1=0$ and $u_{n+1}=\frac{u_n+1}{u_n+2}$. It is easy to check that $u_n$ stays in $(0,1)$, is increasing and converges to $\frac{-1+\sqrt{5}}{2}$.

Equally easily, we have $f(B_n)\subseteq A_n$ and $f(A_n)\subseteq B_{n-1}$, whence $r=2n$ whenever $x_0\in A_n$ and $r=2n+1$ whenever $x_0\in B_n$.

So the only case that's left is when $x_0$ is not in any of the $A_n$ or $B_n$. This means that $x_0$ is at one of the endpoints of $A_n,B_n (n\geq 1)$, i.e. $x_0$ is either one of your $a_k$'s or is $\frac 1 2 - \frac {\sqrt 5}2$.

Edit by the OP (ahorn):

We have seen from Ewan's answer that we can consider $x_0>0$ without loss of generality. What follows is an attempt to solidify the claim that $(x_n)$ converges in this case. Taking advice from Ewan, let $g=f\circ f$ where $f(x):=1+\frac 1 x$. That is, $g(x)=2-\frac 1{x+1}$ which is an increasing function.

Let $x_0\in(0,\phi)$, where $\phi=\frac 1 2 +\frac{\sqrt 5} 2$.

Since $x<g(x)<\phi=\sup\{g(x)\ |\ x\in(0,\phi) \}$ when $x\in(0,\phi)$, $$ x_{2n}<g(x_{2n})=x_{2n+2}<\phi=\sup\{x_{2k}\} $$ where $n, k\in \Bbb N$. So $(x_{2n})$ is an increasing sequence that converges to $\phi$.

$x_0\in(0,\phi)\implies x_{1}\in(\phi,\infty)$.

Since $x>g(x)>\phi=\inf\{g(x)\ |\ x\in(\phi,\infty) \}$ when $x\in(\phi, \infty)$, $$ x_{2n+1}>g(x_{2n+1})=x_{2n+3}>\phi=\inf\{x_{2k+1}\} $$ so $(x_{2n+1})$ is a decreasing sequence that converges to $\phi$.

Now, $x_n\in(0,\phi)\implies x_{n+1}\in(\phi,\infty)$ and $x_n\in(\phi, \infty)\implies x_{n+1}\in(0, \phi)$, so $x_0\in(0,\phi)$ was chosen without loss of generality.

Suppose that for any $\epsilon>0$, $2k\geq N_1\implies|x_{2k}-\phi|<\epsilon$ and $2k+1\geq N_2\implies|x_{2k+1}-\phi|<\epsilon$. Let $N=\max\{N_1, N_2\}$ so that $n\geq N \implies |x_n-\phi|<\epsilon.$

Answer 3

Assuming $f(x)=1+\frac{1}{x}$...

Part 1. First of all (as it was mentioned in a comment), we can't have $x_0=0$. We also can't have $x_0=-1$, because this leads to $x_1=0$. Same applies to $x_0=-\frac{1}{2}$, leading to $x_1=-1$ and $x_2=0$. By induction this applies to any $x_0=-\frac{F_{n-1}}{F_n}$, where $\{F_n\}$-Fibonacci numbers, because $f\left(-\frac{F_{n-1}}{F_n}\right)=1-\frac{F_{n}}{F_{n-1}}=-\frac{F_{n-2}}{F_{n-1}}$ eventually leading to $0$.

Part 2. Let's consider $g(x)=x-f(x)$ which is continuous, except $x=0$, and $g\left(\frac{3}{2} \right)=-\frac{1}{6}<0$ and $g\left(2\right)=\frac{1}{2}>0$. Then $\exists x_0 \in \left(\frac{3}{2},2\right): g(x_0)=0$ or $f(x_0)=x_0$. Additionally, $g'(x)=1+\frac{1}{x^2}>0$ for $x>1$ (i.e. ascending), so that $x_0$ is also unique. We can conclude that if $\lim_{n\rightarrow \infty} x_n$ exists on $\left(1,+\infty\right)$ then it's unique and between $\left(\frac{3}{2},2\right)$. But does it exist?

Part 3. We observe that $\left|f'(x)\right|=\left|\frac{1}{x^2}\right|<1,x>1$ - making $f$ a contraction on $\left[\frac{3}{2},2 \right]$, from the perspective of mean value theorem, i.e. $\exists c \in (x_{n}, x_{n+1})$ such that $|f(x_{n+1}) - f(x_{n})|=|f'(c)||x_{n+1}-x_{n}|$ or $|x_{n+2} - x_{n+1}| < \left|f'(\frac{3}{2})\right| |x_{n+1}-x_{n}|=\frac{4}{9} |x_{n+1}-x_{n}|$. Or $$|x_{n+2} - x_{n+1}| < \left(\frac{4}{9}\right)^2|x_{n}-x_{n-1}| < ... <\left(\frac{4}{9}\right)^n|x_{1}-x_{0}|$$

and $\frac{3}{2}\leq x_0\leq 2 \Rightarrow \frac{2}{3}\geq \frac{1}{x_0} \geq \frac{1}{2} \Rightarrow 2 > \frac{5}{3} \geq f(x_0)\geq \frac{3}{2}$. This can be used to show the limit exists.

Special consideration for the case $x_0 > 2$. Then

$1 < f(x_0)<\frac{3}{2}$, $2>f(f(x_0))>\frac{5}{3}$, $\frac{3}{2}<f^{(3)}(x_0)<\frac{8}{5}$, $\frac{5}{3}>f^{(4)}(x_0)>\frac{13}{8}$ ... it is squeezed in between fractions of form $\frac{F_{n+1}}{F_n}$ which leads to golden ratio.

Part 4. If we take $\forall x_0 \in \left(0,\frac{3}{2}\right)$ we have $x_1=f(x_0)>\frac{2}{3}+1>\frac{3}{2}$. From $x_1$ the sequence "falls" in the convergence zone $\left[\frac{3}{2},+\infty \right)$.

Part 5. $\forall x_0 \in \left(-\infty,-1\right)$ we have $x_0<-1 \Rightarrow -x_0>1 \Rightarrow 0< -\frac{1}{x_0} <1 \Rightarrow 0>\frac{1}{x_0} > -1 \Rightarrow x_1=f(x_0) > 0$. According to Part 3 & 4, either from $x_1$ or $x_2$ the sequence "falls" in the convergence zone $\left[\frac{3}{2},+\infty \right)$

Part 6. $\forall x_0 \in \left(-\frac{1}{2},0\right)$ we have $-\frac{1}{2} < x_0 <0 \Rightarrow \frac{1}{2} > -x_0 >0 \Rightarrow 2 < -\frac{1}{x_0} \Rightarrow -2 > \frac{1}{x_0} \Rightarrow x_1=f(x_0) < -1$ from $x_1$ we are in the Part 5 scenario.

Part 7. This time we will have to look in between $-1 < -\frac{2}{3} < -\frac{5}{8}<...<-\frac{F_{n-1}}{F_n}<...<-\frac{8}{13}< -\frac{3}{5}< -\frac{1}{2}$, considering the alternating nature of $\frac{F_{n-1}}{F_n}$, we observe that:

• $x_0 \in \left(-1, -\frac{2}{3}\right) \Rightarrow f(x_0) \in \left(-\frac{1}{2},0\right)$, this is Part 6.
• $x_0 \in \left(-\frac{3}{5}, -\frac{1}{2}\right) \Rightarrow f(x_0) \in \left(-1, -\frac{2}{3}\right)$ this is previous case now.
• ... and so on, every $A$ gets promoted to $f(A)$ previously analysed escaping eventually into convergence zone, just an example $$-\frac{F_{n-1}}{F_{n}} < x_0 < -\frac{F_{n+1}}{F_{n+2}} \Rightarrow \frac{F_{n-1}}{F_{n}} > -x_0 > \frac{F_{n+1}}{F_{n+2}} \Rightarrow \frac{F_{n}}{F_{n-1}} < -\frac{1}{x_0} < \frac{F_{n+2}}{F_{n+1}} \Rightarrow$$ $$1-\frac{F_{n}}{F_{n-1}} > 1 + \frac{1}{x_0} > 1-\frac{F_{n+2}}{F_{n+1}} \Rightarrow -\frac{F_{n-2}}{F_{n-1}} > f(x_0) > -\frac{F_{n}}{F_{n+1}}$$

The only problem is $\lim_{n\rightarrow \infty} -\frac{F_{n-1}}{F_n}$ which is $$\lim_{n\rightarrow \infty} -\frac{F_{n-1}}{F_n}=-\frac{1}{\varphi }=\frac{1-\sqrt{5}}{2}$$

For all the cases, the limit is the positive root of polynomial $x^2-x-1=0$.

Answer 4

Hint: show that the sequence $x_n$ is bounded. It must then have a convergent subsequence. Can you show the whole sequence converges to the same limit?

Answer 5

Define $$a_n=\frac1{x_n-\phi}$$ where $$\phi=\frac{1+\sqrt5}{2}\quad\quad\left(\phi^2-\phi-1=0\right)$$ Then \begin{align} &x_{n+1}=\frac1{a_{n+1}}+\phi=1+\frac1{x_n}=1+\frac1{\frac1{a_{n}}+\phi}\\\\ &\left(\frac1{a_{n+1}}+\phi\right)\left(\frac1{a_{n}}+\phi\right)=\frac1{a_{n}}+\phi+1\\\\ &\frac1{a_{n+1}}\frac1{a_{n}}+\frac{\phi}{a_{n+1}}+\frac{\phi-1}{a_{n}}=0\\\\ &(\phi-1)a_{n+1} + \phi a_n + 1=0\\\\ &(\phi-1)\lambda^2 + \phi \lambda + 1=0\quad\Rightarrow\quad\lambda=-1\,,-\phi\\\\ &a_n=\alpha(-1)^n+\beta(-\phi)^n\\\\ &a_0=\alpha+\beta=\frac1{x_0-\phi}\\\\ &a_1=-\alpha-\phi\beta=\frac1{1+\frac1{x_0}-\phi}=\frac{x_0}{x_0(1-\phi)+1}=\frac{\frac{x_0}{1-\phi}}{x_0-\phi}\\\\ &\beta=\frac1{(\phi-1)^2}\left(1+\frac1{x_0-\phi}\right)\\\\ &\alpha=\frac1{(\phi-1)^2}\left(\frac{1-x_0}{x_0-\phi}\right)\\\\ \end{align} Therefore $$\boxed{\large{\quad x_n=\frac{(\phi-1)^2(x_0-\phi)}{(1-x_0)(-1)^n+(x_0-\phi+1)(-\phi)^n}+\phi\quad}}\\$$ except some special cases such as

$x_0=-\frac1\phi$ where $x_n = x_0$, or

$x_0=-\frac12, -\frac23, -\frac35, -\frac58, \dots$ where $x_n$ hits $0$.

But you can see in most cases $x_n$ converges to $\phi$ because $|-\phi|>1$.