Let \(x_0 = a, x_1 = b\), and set \[ x_n \definedas {x_{n-1} + x_{n-2} \over 2} \quad n\geq 2 .\]
Show that \(\theset{x_n}\) is a Cauchy sequence and find its limit in terms of \(a\) and \(b\).
Solution
The recurrence \[ x_n=\frac{x_{n-1}+x_{n-2}}2 \] has characteristic equation \[ 2r^2-r-1=0, \] whose roots are \(1\) and \(-1/2\). Hence \[ x_n=A+B\left(-\frac12\right)^n. \] The initial conditions give \[ a=A+B, \qquad b=A-\frac B2. \] Solving, \[ A=\frac{a+2b}{3}, \qquad B=\frac{2(a-b)}{3}. \] Therefore \[ x_n=\frac{a+2b}{3} +\frac{2(a-b)}{3}\left(-\frac12\right)^n. \] Since the second term tends to \(0\), \[ \boxed{\lim_{n\to\infty}x_n=\frac{a+2b}{3}}. \] Thus \((x_n)\) converges, and every convergent real sequence is Cauchy.