Define a set of positive real numbers as follows. Let \(x _ { 0 } > 0\) be any positive number, and let \(x _ { n + 1 } = ( 1 + x _ { n } ) ^ { - 1 }\) for all \(n \geq 0\) . Prove that this sequence converges, and find its limit.
Solution
Let \[ f(x)\coloneqq\frac1{1+x}, \] so that \(x_{n+1}=f(x_n)\).
1. For every positive initial value \(x_0\),\[ \frac12<x_n<\frac23 \qquad (n\ge3). \]
Since \(x_0>0\), \[ 0<x_1<1. \] Therefore \[ \frac12<x_2<1, \] and hence \[ \frac12<x_3<\frac23. \] If \(1/2<x<2/3\), then \[ \frac35<f(x)<\frac23, \] so in particular \(1/2<f(x)<2/3\). Induction now gives the displayed bound for every \(n\ge3\).
2. The map \(f\) has a unique positive fixed point\[ \alpha\coloneqq\frac{\sqrt5-1}{2}, \]and \(\alpha\in(1/2,2/3)\).
The fixed-point equation is \[ x=\frac1{1+x}, \] equivalently \[ x^2+x-1=0. \] Its unique positive root is \[ \alpha=\frac{\sqrt5-1}{2}. \] If the iteration starts at \(x_0=\alpha\), then every term equals \(\alpha\). Applying step 1 to this particular initial value gives \(1/2<\alpha<2/3\).
3. On \((1/2,2/3)\) one has\[ \lvert f'(x)\rvert<\frac49. \]
Since \[ f'(x)=-\frac1{(1+x)^2}, \] the inequality \(x>1/2\) gives \[ \lvert f'(x)\rvert =\frac1{(1+x)^2} <\frac1{(3/2)^2} =\frac49. \]
4. For every \(n\ge3\),\[ \lvert x_{n+1}-\alpha\rvert \le \frac49\lvert x_n-\alpha\rvert. \]
By steps 1 and 2, both \(x_n\) and \(\alpha\) lie in \((1/2,2/3)\). The mean value theorem and step 3 give \[ \lvert f(x_n)-f(\alpha)\rvert \le \frac49\lvert x_n-\alpha\rvert. \] Since \(f(x_n)=x_{n+1}\) and \(f(\alpha)=\alpha\), this is the desired estimate.
5. The sequence converges to\[ \boxed{\frac{\sqrt5-1}{2}}. \]
Iterating step 4 gives, for \(n\ge3\), \[ \lvert x_n-\alpha\rvert \le \left(\frac49\right)^{n-3} \lvert x_3-\alpha\rvert. \] The right-hand side tends to \(0\), so \(x_n\to\alpha\).
6. Q.E.D.
Step 5 proves convergence and identifies the limit.