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I'm doing Problem II.3.4 in textbook Analysis I by Amann/Escher.

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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.

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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 $x_n \to 1$.

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  • $\begingroup$ Your proof is much more elegant. Thank you so much ;) $\endgroup$
    – Akira
    Aug 21 '19 at 5:36
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Just a hint:

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 $$y_{n+1}= y_n^2$$

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