The original problem is from Zorich Mathematical Analysis I 3.1 Exercises:
a) Show that if $a>0$, the sequence $x_{n+1}=\frac{1}{2}(x_n+\frac{a}{x_n})$ converges to the squareroot of $a$ for any $x_1>0$
I have managed to do this part, by monotone convergence theorem. But couldn't do part b) of this problem:
b) Estimate the rate of convergence, that is, the magnitude of absolute error $|x_n-\sqrt{a}|=|\Delta_n|$as a function of n.
My attempt of a):Let $x_{n+1}=\frac{1}{2}(x_n+\frac{a}{x_n})$Notice that
$$x_n-x_{n+1}=x_n-\frac12\Bigl(x_n+\frac a{x_n}\Bigr)=\frac1{2x_n}(x_n^2-a).$$In use of the previous result, we get:$$\begin{align*}x_n^2-a&=\frac14\Bigl(x_{n-1}+\frac a{x_{n-1}}\Bigr)^2-a\\&=\frac{x_{n-1}^2}4-\frac a2+\frac{a^2}{4x_{n-1}^2}\\&=\frac14\Bigl(x_{n-1}^2-2a+\frac{a^2}{x_{n-1}^2}\Bigr)\\&=\frac{1}{4}\Bigl(x_{n-1}-\frac a{x_{n-1}}\Bigr)^2\\&\ge0.\end{align*}$$Therefore, $x_n\ge x_{n+1}$ and that $x_n$ is bounded below，because $x_n^2\ge a$ for each $n\ge2$.
Due to monotone convergence theorem, we have that a sequence that is bounded below and monotonely decreasing(strictly or unstrictly) has a limit.
We denote the limit as $x=\lim_{n\to\infty}x_n$. Therefore we have:$$x=\lim_{n\to\infty}x_n\quad\iff x=\frac{1}{2}(x+\frac ax)\implies\quad x=\sqrt a.$$
My attempt of b) so far:
$$\Delta_{n+1} = \left| x_{n+1} - \sqrt{a} \right| \quad,\Delta_{n+1} = \left| \frac{1}{2}\left( x_{n}+\frac{a}{x_{n}} \right) - \sqrt{a} \right|\Delta_{n}$$$$\Delta_{n+1}=\frac{1}{2}\left|\left(\frac{a}{x_{n}}-\sqrt{ a }+x_{n} -\sqrt{a }\right)\right|\Delta_{n}$$$$\Delta_{n+1}=\frac{1}{2}\left|\frac{a-\sqrt{ a}x_{n}}{x_{n}}+x_{n}-\sqrt{ a }\right|\Delta_{n}$$$$\Delta_{n+1}=\frac{1}{2}\left|\sqrt{a }\left(\frac{ \sqrt{ a }-x_{n}}{x_{n}}+x_{n}-\sqrt{ a }\right)\right|\Delta_{n}$$Since that either $x_{n}\ge \sqrt{ a}\ge {0}$ or $x_{n}\le\sqrt{ a }$, and that $|a|=|-a|$, and that $x_{n}\ge a$ for each $n\ge 2$We have:
$$\Delta_{n+1}=\frac{1}{2} \bigg|\sqrt{ a } \frac{\Delta_{n}}{x_{n}}-\Delta_{n}\bigg|$$$$\implies{2}\Delta_{n+1}=\Delta_{n}|\frac{\sqrt{ a }-x_{n}}{x_{n}}|$$$$\implies 2\Delta_{n+1}=\Delta^2_{n}|\frac{1}{x_{n}}|$$$$\implies \Delta_{n+1}=\frac{1}{2}\Delta^2_{n}|\frac{1}{x_n}|$$
We can then calculate with $\Delta_{1}=|x-\sqrt{ a } |$, and the result we got above$x_{n+1}=\frac{1}{2}(x_n+\frac{a}{x_n})$