Fix a>0 and let $\displaystyle x_1>\sqrt{a}$. For n >=1, define

$\displaystyle x_{n+1}$ = $\displaystyle 1/2(x_n +a/x_n)$

Show that $\displaystyle (x_n)$ converges and that lim n--> infinity $\displaystyle x_n = \sqrt{a}$.

Here is what I have got so far, $\displaystyle x_{n+1}$= $\displaystyle 1/2(x_n +a/x_n)$ $\displaystyle >= \sqrt{a}$ since $\displaystyle \sqrt{ab}<=(a+b)/2$