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:
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Basis step: $x_1=2, x_1^2 = 4 >2$.
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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!!
Best Answer
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 $a=\sqrt{2}$.