Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$ 
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.

$\bullet~$ I can show that $x^2_n \geqslant 2$ by induction:

*

*Basis step: $x_1=2, x_1^2 = 4 >2$.


*Inductive step:
Assume we have some $k$ for which $x^2_K \geqslant 2$, then
\begin{align*}
(x_{k+1})^2 =&~ \bigg(\dfrac{1} {2} \bigg(x_k + \dfrac{2} {x_k}\bigg)\bigg)^2\\
=&~ \dfrac{1} {4} x_k^2 + 1 + \dfrac{1} {x_k^2}\\
\geqslant&~ \dfrac {1} {4} x_k^2 + 1 \geqslant 2\\
&~\dfrac {1} {4} x_k^2 \geqslant 1  \quad \forall~ n \geqslant 2
\end{align*}
How do I use this to show that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$
and hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$?
Thanks!!
 A: Note the fact that $x_{n+1} =\frac{1}{2}(x_n+\frac{2}{x_{n}}) \le \frac{1}{2}(x_n+x_{n})=x_{n}$ from $x_{n}^2 \ge 2$ (or $x_{n} \ge \frac{2}{x_{n}}$)
Thus, $x_{n}-x_{n+1}\le0$. 
This implies that $x_{n}$ is a decreasing sequence that is greater than $\sqrt{2}$, implying that $x_{n}$ is convergent. 
Let $\lim{x_n}=a$.
Note the fact that $a=\frac{1}{2}(a+\frac{2}{a})$. This implies that $a=\sqrt{2}$. 
A: $$x_n-x_{n+1}=x_n-\frac12\left(x_n+\frac 2{x_n}\right)=\frac{2x_n^2-x_n^2-2}{2x_n^2}=\frac{x_n^2-2}{2x_n^2}.$$
A: By $AM\geq GM$, 
$x_{n+1}=\frac{1}{2} \left( x_{n}+\frac{2}{x_{n}} \right)
\geq \sqrt{2}$ and $x_{1}=2$, therefore
$$x_{n}^{2} \geq 2 \: , \quad \forall n\in \mathbb{N}$$
Now,
\begin{align*}
  x_{n}-x_{n+1} &=\frac{x_{n}^{2}-2}{x_{n}} \\
                &\geq 0
\end{align*}
So $\{ x_{n} \}$ is decreasing and
\begin{align*}
  x_{n}-x_{n+1} &\leq \frac{x_{1}^{2}-2}{2^{n/2}} \\
                &= \frac{2}{2^{n/2}}
\end{align*}
Hence $\lim_{n\to \infty} (x_{n}-x_{n+1})=0$ and $\{ x_{n} \}$ is decreasing but bounded below, 
$L=\lim_{n\to \infty} x_{n}$ exists.
Therefore $L=\frac{1}{2} \left( L+\frac{2}{L} \right) \implies L=\sqrt{2}$
