Here is Prob. 16, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix a positive number $\alpha$. Choose $x_1 > \sqrt{\alpha}$, and define $x_2, x_3, x_4, \ldots$, by the recursion formula $$ x_{n+1} = \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right). $$
(a0 Prove that $\left\{ x_n \right\}$ decreases monotonically and that $\lim x_n = \sqrt{\alpha}$.
(b) Put $\epsilon_n = x_n - \sqrt{\alpha}$, and show that $$ \epsilon_{n+1} = \frac{\epsilon_n^2}{2x_n} < \frac{\epsilon_n^2}{2\sqrt{\alpha}}$$ so that, setting $\beta = 2 \sqrt{\alpha}$, $$ \epsilon_{n+1} < \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^n} \qquad (n = 1, 2, 3, \ldots). $$
(c) This is a good algorithm for computing square roots, since the recursion formula is simple and the convergence is extremely rapid. For example, if $\alpha = 3$ and $x_1 = 2$, show that $\epsilon_1/\beta < \frac{1}{10}$ and that therefore $$ \epsilon_5 < 4 \cdot 10^{-16}, \qquad \epsilon_6 < 4 \cdot 10^{-22}.$$
My Attempt:
Part (a):
As $x_1 > \sqrt{\alpha} > 0$, so $x_1^2 > \alpha > 0$, and we have $$ x_1 - x_2 = x_1 - \frac{1}{2} \left( x_1 + \frac{\alpha}{x_1} \right) = \frac{1}{2} \left( x_1 - \frac{\alpha}{x_1} \right) = \frac{ x_1^2 - \alpha }{2 x_1} > 0, $$ which implies that $$x_1 > x_2. \tag{1} $$
And, $$ x_2 - \sqrt{\alpha} = \frac{1}{2} \left( x_1 + \frac{\alpha}{x_1} \right) - \sqrt{\alpha} = \frac{1}{2} \left( \sqrt{x_1} - \frac{\sqrt{\alpha}}{\sqrt{x_1}} \right)^2 > 0,$$ which implies that $$ x_2 > \sqrt{\alpha}. \tag{2} $$
Suppose that, for some natural number $n \geq 2$, $x_n > \sqrt{\alpha} > 0$. Then we have $x_n^2 > \alpha > 0$, and so $$ x_n - x_{n+1} = x_n - \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right) = \frac{1}{2} \left( x_n - \frac{\alpha}{x_n} \right) = \frac{x_n^2 - \alpha}{2x_n} > 0,$$ which implies that $x_n > x_{n+1}$.
And, $$x_{n+1} - \sqrt{\alpha} = \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right) - \sqrt{\alpha} = \frac{1}{2} \left( x_n - 2 \sqrt{\alpha} + \frac{\alpha}{x_n} \right) = \frac{1}{2} \left( \sqrt{x_n} - \frac{\sqrt{\alpha}}{\sqrt{x_n}} \right)^2 > 0,$$ which implies that $x_{n+1} > \sqrt{\alpha}$.
Therefore, we have $$x_1 > x_2 > \cdots > \sqrt{\alpha},$$ and so $\lim x_n$ exists. Let $x = \lim x_n$. Then $x \geq \sqrt{\alpha} > 0$ and $$ x = \frac{1}{2} \left( x + \frac{\alpha}{x} \right),$$ which implies that $$ x^2 = \alpha. $$ Hence $\lim x_n = \sqrt{\alpha}$, as required.
Is this proof correct and the same as intended by Rudin?
Part (b):
We note that $$ \begin{align} \epsilon_{n+1} &= x_{n+1} - \sqrt{\alpha} \\ &= \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right) - \sqrt{\alpha} \\ &= \frac{x_n^2 - 2x_n \sqrt{\alpha} + \alpha }{2x_n} \\ &= \frac{ (x_n - \sqrt{\alpha} )^2}{2x_n} \\ &= \frac{\epsilon_n^2}{2x_n} \\ &< \frac{\epsilon_n^2}{2\sqrt{\alpha}}. \end{align} $$ Setting $\beta = 2 \sqrt{\alpha}$, we find that $$\epsilon_{n+1} < \frac{\epsilon_n^2}{\beta}. \tag{3} $$ Thus $$\epsilon_2 < \frac{\epsilon_1^2}{\beta} = \beta \left( \frac{\epsilon_1}{\beta} \right)^2 = \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^1}. $$ Now suppose that for some natural number $n$, we have $$ \epsilon_{n+1} < \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^n}.$$ Then from (3), we see that $$ \epsilon_{n+2} < \frac{\epsilon_{n+1}^2}{\beta} = \beta \left( \frac{\epsilon_{n+1} }{\beta} \right)^2 < \beta \left( \frac{ \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^n} }{ \beta } \right)^2 = \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^{n+1}}, $$ and the required result follows by induction.
Is my proof correct and as expected by Rudin?
Part (c):
If $\alpha = 3$ and $x_1=2$, then $\beta = 2 \sqrt{\alpha} = 2 \sqrt{3}$, and $\epsilon_1 = 2-\sqrt{3}$; so $$\frac{\epsilon_1}{\beta} = \frac{2-\sqrt{3}}{2\sqrt{3}} = \frac{1}{2 \sqrt{3} \left( 2+ \sqrt{3} \right) } = \frac{1}{6+4 \sqrt{3} } < \frac{1}{6 + 4} = \frac{1}{10} $$ because $1 < \sqrt{3}$.
So $$ \epsilon_5 < \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^4} < 4 \left( \frac{1}{10} \right)^{16} = 4 \cdot 10^{-16},$$ and $$\epsilon_6 < \beta \left( \frac{\epsilon_1}{\beta} \right)^{2^5} < 4 \left( \frac{1}{10} \right)^{32} = 4 \cdot 10^{-32}, $$ as required.
Are my calculations correct?
