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, 


Observe that
$$\frac{x_n-\sqrt2}{x_n+\sqrt2}=\frac{\dfrac{x_{n-1}}{2} + \dfrac{1}{x_{n-1}}-\sqrt2}{\dfrac{x_{n-1}}{2} + \dfrac{1}{x_{n-1}}+\sqrt2}=\frac{x_{n-1}^2-2\sqrt2x_{n-1}+2}{x_{n-1}^2+2\sqrt2x_{n-1}+2}=\left(\frac{x_{n-1}-\sqrt2}{x_{n-1}+\sqrt2}\right)^2.$$
Then by induction,
$$\frac{x_n-\sqrt2}{x_n+\sqrt2}=\left(\frac{x_0-\sqrt2}{x_0+\sqrt2}\right)^{2^n}.$$
Then whenever
$$\left|\frac{x_0-\sqrt2}{x_0+\sqrt2}\right|<1$$ (in particular $x_0>0$) the series converges to $\sqrt2$.
