Prove $(x_n)$ defined by $x_n= \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}}$ converges when $x_0>1$

Question

$x_n= \dfrac{x_{n-1}}{2} + \dfrac{1}{x_{n-1}}$

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,

Answer 1

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 $\sqrt{2}$.

Answer 2

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 $n \ge 1$.

Answer 3

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 $\sqrt2$.