Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$ 
Let $x_1 = 2$, and define $$x_{n+1} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right).$$ Show that $x_n^2$ is always greater than or equal to $2$, and then use this to prove that $x_n − x_{n+1} ≥ 0$. Conclude that $\lim x_n =
\sqrt2$.

My question:
So I know how to do this problem but I don't know how to prove that $x_n >0$ for all $n\in \mathbb{N}$.

From
$x_{n+1} 
= \frac{1}{2} \left(x_n + \frac{2}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2 
= \frac{1}{4} \left(x_n^2 +4+ \frac{4}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -2
&= \frac{1}{4} \left(x_n^2 -4+ \frac{4}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{2}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-2}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{2}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).

More generally,
using copy, paste, and edit:
From
$x_{n+1} 
= \frac{1}{2} \left(x_n + \frac{a}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2 
= \frac{1}{4} \left(x_n^2 +2a+ \frac{a^2}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -a
&= \frac{1}{4} \left(x_n^2 -2a+ \frac{a^2}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{a}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-a}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{a}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).
