# For $x_0 \ge 1$, the sequence $(x_n)$ defined recursively by $x_{n+1} = (x_n +1/x_n)/2$ converges to $1$

I'm doing Problem II.3.4 in textbook Analysis I by Amann/Escher. After elementary transformations, the problem is equivalent to the below theorem:

Theorem: For $$x_0 \ge 1$$, the sequence $$(x_n)$$ defined recursively by $$x_{n+1} = (x_n +1/x_n)/2$$ converges to $$1$$.

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

First, we prove that this sequence is convergent. By AM-GM inequality, $$x_{n+1} = (x_n +1/x_n)/2 \ge 1$$ for all $$n$$, so the sequence is bounded from below. We have $$x_{n+1} - x_n = (1-x_n^2)/(2x_n) \le 0$$ and thus the sequence is decreasing. As such, $$\lim_{n \to \infty} x_n =a \in \mathbb R^+$$.

Next, we prove that $$a=1$$. We have

\begin{aligned}a &= \lim_{n \to \infty} x_n &&= \lim_{n \to \infty} x_{n+1} \\ &= \lim_{n \to \infty} (x_n +1/x_n)/2 &&= \left ( \lim_{n \to \infty} x_n + \dfrac{1}{\lim_{n \to \infty} x_n} \right)/2 \\ &=(a+1/a)/2 \end{aligned}

This equation implies $$a=1$$. This completes the proof.

Your proof is fine. You showed that the sequence is decreasing and bounded below (thus convergent). You might elaborate that the equation $$a = (a+1/a)/2$$ has two solutions ($$a= \pm 1$$), but only $$a=1$$ can be the limit.
Alternatively observe that $$0 \le x_{n+1} - 1 = \frac{(x_n-1)^2}{2x_n} \le \frac{(x_n-1)^2}{2}$$ which also implies convergence $$x_n \to 1$$.
With $$y = \frac{1}{2}( x + a/x)$$ we have $$y \pm \sqrt{a} = \frac{(x\pm \sqrt{a})^2}{2 x}$$ so $$\frac{y-\sqrt{a}}{y+\sqrt{a}}= \left(\frac{x-\sqrt{a}}{x+\sqrt{a}}\right)^2$$
Now we are reduced to investigating the recurrence $$y_{n+1}= y_n^2$$