Take \(x_0 = a\), \(x_1 = b\), and set \(x_n \coloneqq \frac{x_{n-1} + x_{n-2}}{2}\) for \(n \geq 2\). Prove that \((x_n)\) is a Cauchy sequence and find its limit in terms of \(a\) and \(b\).
Solution
For \(n\geq1\), put \[ d_n=x_n-x_{n-1}. \]
1. For every \(n\geq2\),\[ d_n=-\frac12d_{n-1}, \]and hence\[ d_n= \left(-\frac12\right)^{n-1}(b-a). \]
Using the recurrence for \(x_n\), \[ \begin{aligned} d_n &=x_n-x_{n-1}\\ &=\frac{x_{n-1}+x_{n-2}}2-x_{n-1}\\ &=-\frac12(x_{n-1}-x_{n-2})\\ &=-\frac12d_{n-1}. \end{aligned} \] Since \[ d_1=x_1-x_0=b-a, \] iteration gives the stated formula.
2. For every \(n\geq0\),\[ x_n = \frac{a+2b}{3} + \frac{2(a-b)}3 \left(-\frac12\right)^n. \]
For \(n\geq1\), telescoping and step 1 give \[ \begin{aligned} x_n &=a+\sum_{k=1}^n d_k\\ &=a+(b-a)\sum_{k=1}^n\left(-\frac12\right)^{k-1}\\ &=a+\frac23(b-a) \left(1-\left(-\frac12\right)^n\right)\\ &=\frac{a+2b}{3} +\frac{2(a-b)}3 \left(-\frac12\right)^n. \end{aligned} \] For \(n=0\) the same expression equals \(a\), so the formula holds for all \(n\geq0\).
3. The sequence \((x_n)\) is Cauchy.
Put \[ C=\frac{2(a-b)}3. \] By step 2, for any \(m,n\geq0\), \[ \begin{aligned} \abs{x_n-x_m} &= \abs{C} \abs{ \left(-\frac12\right)^n - \left(-\frac12\right)^m }\\ &\leq \abs{C} \left(2^{-n}+2^{-m}\right). \end{aligned} \] Let \(\varepsilon>0\). If \(C=0\), then the sequence is constant. Otherwise choose \(N\) so large that \[ 2\abs{C}\,2^{-N}<\varepsilon. \] For \(m,n\geq N\), the displayed estimate gives \[ \abs{x_n-x_m}<\varepsilon. \] Thus \((x_n)\) is Cauchy.
4. The limit is\[ \boxed{ \lim_{n\to\infty}x_n = \frac{a+2b}{3} }. \]
By step 2, \[ x_n-\frac{a+2b}{3} = \frac{2(a-b)}3 \left(-\frac12\right)^n. \] Since \[ \left(-\frac12\right)^n\longrightarrow0, \] the displayed difference tends to zero.
5. Q.E.D.
Step 3 proves the Cauchy property, and step 4 gives the requested limit.