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, 


We have: By AM-GM inequality: $x_n > 2\sqrt{\dfrac{x_{n-1}}{2}\cdot \dfrac{1}{x_{n-1}}}=\sqrt{2}, \forall n \geq 1$. Thus: $x_n-x_{n-1} = \dfrac{1}{x_{n-1}} - \dfrac{x_{n-1}}{2}= \dfrac{2-x_{n-1}^2}{2x_{n-1}}  < 0$. Hence $x_n$ is a decreasing sequence,and is bounded below by $\sqrt{2}$. So it converges, and you showed the limit is $\sqrt{2}$.
