Let $a > 0$. Choose $x_1 >$ $\sqrt a$. Define a sequence {$x_n$} recursively as $x_{n+1} = 1/2(x_n + a/x_n)$ for $n > 1$.
Prove that lim$x_n = \sqrt a$.
I think I first want to prove that {$x_n$} is bounded below by $\sqrt a$. But, I am having trouble with this. Could someone help me out a little?
We have $$x_{n+1}-\sqrt{a}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)-\sqrt{a}=\frac{1}{2}\left(x_n -2\sqrt{a}+\frac{a}{x_n}\right).\tag{1}$$
Note that the expression on the right of (1) is equal to $$\frac{1}{2}\left(\sqrt{x_n} -\frac{\sqrt{a}}{\sqrt{x_n}}\right)^2.$$ Since it is a square, it is $\ge 0$.
Remark: In the calculation, we assumed that $x_n$ is always $\gt 0$. This is obvious, since $x_1\gt 0$. If we want to be very formal, we can prove that $x_n\gt 0$ by induction.
Here is an unorthodox approach :-). Apply Newton's method to $f(x) = x^2 - a$ and you will find exactly your iteration: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^2-a}{2x_n} = \frac{x_n^2+a}{2x_n} $$ thus you are just applying Newton's method to find the roots of $f(x)$ starting from $x_1 > \sqrt{a}$. As is easy to see, the roots are $x = \pm \sqrt{a}$ and the closest root is $\sqrt{a}$, where Newton will converge.
