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
Put \(d_n=x_n-x_{n-1}\). Then \[ d_n=-{1\over2}d_{n-1}, \qquad |d_n|=2^{-(n-1)}|b-a|. \] If \(m>n\), then \[ |x_m-x_n| \le\sum_{j=n+1}^m|d_j| \le |b-a|\sum_{j=n+1}^\infty2^{-(j-1)} =2^{1-n}|b-a|, \] which tends to \(0\) as \(n\to\infty\). Thus \((x_n)\) is Cauchy.
To identify the limit, observe that \[ x_{n-1}+2x_n =x_{n-1}+x_{n-1}+x_{n-2} =x_{n-2}+2x_{n-1}. \] Hence this quantity is independent of \(n\), and from \(n=1\) it equals \[ a+2b. \] Since the sequence is Cauchy in \(\CC\), let \(x_n\to L\). Passing to the limit in the invariant gives \[ 3L=a+2b, \qquad L={a+2b\over3}. \]