Prove $(x_n)$ defined by $x_n= \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}}$ converges when $x_0>1$ $x_n= \dfrac{x_{n-1}}{2} + \dfrac{1}{x_{n-1}}$
I know it converges to $\sqrt2$ and I do not want the answer. I just want a prod in the right direction. 
I have tried the following and none have worked:  $x_n-x_{n-1}$ and this got me nowhere.
I have tried $\dfrac{x_{n+1}}{x_{n}}$ with no luck, and I was expecting this  to help, 


Let $f(x)=x^2-2$, note that $f$ is convex and hence it is straightforward to show that if $x_n$ is the Newton iteration for the equation $f(x) = 0$, and 
$x_0>0$, then $x_n \ge \sqrt{2}$ (equivalently, $f(x_n) \ge 0$) if $n \ge 1$, and $\sqrt{2} \le x_{n+1} \le x_n$ for $n \ge 1$.
