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!!

$$x_n-x_{n+1}=x_n-\frac12\left(x_n+\frac 2{x_n}\right)=\frac{2x_n^2-x_n^2-2}{2x_n^2}=\frac{x_n^2-2}{2x_n^2}.$$
