Proving a sequence converges by defining the sequence recursively.

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?

Solution 1:

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.