(c.f. [Abb01, Exercise 2.4.5]) Let \(x_1=2\) and define \[ x_{n+1}=\frac12\left(x_n+\frac{2}{x_n}\right). \] Find \(\lim_{n\to\infty}x_n\). Hint: Show that \(\{x_n\}\) is decreasing.
Solution
1. \(x_n \ge \sqrt2\) for all \(n\).Proof: by AM-GM, \(x_{n+1} = \frac12\big(x_n + \frac{2}{x_n}\big) \ge \sqrt{x_n\cdot\frac{2}{x_n}} = \sqrt2\) for every \(n\); since \(x_1 = 2 \ge \sqrt2\), all \(x_n \ge \sqrt2\).
2. The sequence is decreasing from \(x_2\) onward.Proof: for \(n \ge 1\), \[\frac{x_{n+1}}{x_n} = \frac12\left(1 + \frac{2}{x_n^2}\right) \le \frac12\left(1 + \frac{2}{2}\right) = 1,\] using \(x_n^2 \ge 2\) from 1; so \(x_{n+1} \le x_n\) for all \(n\) (including \(n=1\)). Thus \((x_n)\) is decreasing and bounded below.
3. The sequence converges and we identify the limit.Proof: \((x_n)\) is decreasing and bounded below by \(\sqrt2\), so it converges to some \(L \ge \sqrt2\).Taking limits in the recursion (the map \(x \mapsto \frac12(x + 2/x)\) is continuous on \([\sqrt2, \infty)\)), \[L = \frac12\left(L + \frac{2}{L}\right) \implies 2L = L + \frac{2}{L} \implies L^2 = 2 \implies L = \sqrt2\] (as \(L \ge \sqrt2 > 0\)).
4. Q.E.D.