Show that \[ x_{n+1}=(1+x_n)^{-1} \] converges and find its limit for any \(x_0>0\).
Solution
Let \[ \alpha=\frac{\sqrt5-1}{2}. \]
1. Every term \(x_n\) is positive.
The hypothesis gives \(x_0>0\). If \(x_n>0\), then \[ x_{n+1}=\frac{1}{1+x_n}>0. \] The claim follows by induction.
2. The number \(\alpha\) is the unique positive fixed point of the map\[ T(x)=\frac{1}{1+x}. \]
The fixed-point equation is \[ \alpha=\frac{1}{1+\alpha}, \] equivalently \[ \alpha^2+\alpha-1=0. \] Its two roots are \[ \frac{-1\pm\sqrt5}{2}, \] and exactly one of them is positive, namely \(\alpha=(\sqrt5-1)/2\). In particular, \[ \frac{1}{1+\alpha}=\alpha \qquad\text{and}\qquad 0<\alpha<1. \]
3. For every \(n\geq0\),\[ \abs{x_{n+1}-\alpha} \leq \alpha\abs{x_n-\alpha}. \]
Using step 2, \[ \begin{aligned} \abs{x_{n+1}-\alpha} &= \left\lvert \frac{1}{1+x_n}-\frac{1}{1+\alpha} \right\rvert\\ &= \frac{\abs{x_n-\alpha}}{(1+x_n)(1+\alpha)}. \end{aligned} \] By step 1, \(x_n>0\), so \(1+x_n\geq1\). Therefore \[ \frac{1}{(1+x_n)(1+\alpha)} \leq \frac{1}{1+\alpha} = \alpha, \] where the last equality is step 2.
4. For every \(n\geq0\),\[ \abs{x_n-\alpha} \leq \alpha^n\abs{x_0-\alpha}. \]
3. The case \(n=0\) is equality. If the estimate holds at \(n\), then\[ \abs{x_{n+1}-\alpha} \leq \alpha\abs{x_n-\alpha} \leq \alpha^{n+1}\abs{x_0-\alpha}. \]
5. The sequence converges and\[ \boxed{\displaystyle\lim_{n\to\infty}x_n=\frac{\sqrt5-1}{2}}. \]
By step 2, \(0<\alpha<1\), so \[ \alpha^n\abs{x_0-\alpha}\longrightarrow0. \] Step 4 therefore gives \[ \abs{x_n-\alpha}\longrightarrow0, \] which is exactly \(x_n\to\alpha\).
6. Q.E.D.
Step 5 proves convergence and gives the requested limit.