Given $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $, prove $\sqrt{2} ≤ x_n ≤ 2$

Note that $\sqrt{2} \leq \sqrt{2\sqrt{2}}\leq x_{k+1} \leq 2$ so more generally, we can say that $\sqrt{2} \leq x_{k+1} \leq 2$ because if $\sqrt{2} \leq \sqrt{2\sqrt{2}}$ and $\sqrt{2\sqrt{2}}\leq x_{k+1}$, then by transitivity of partial ordering, $\sqrt{2} \le x_{k+1}$ (What happened to the numbers between $\sqrt{2}$ and $\sqrt{2\sqrt{2}}$? Well, I think we can make this generalization because in this sequence, there is no number between them. This is the result of second part).

From

$$\sqrt2<x_n<2$$ you can deduce

$$\sqrt{2\sqrt2}<\sqrt{2x_n}<2$$

and obviously

$$\sqrt{2}<\sqrt{2x_n}<2,$$ which is nothing but

$$\sqrt{2}<x_{n+1}<2.$$

Hints:

• $\;\displaystyle \frac{x_{n+1}}{2}=\sqrt{\frac{x_{n}}{2}}\,$, so $\;\displaystyle \frac{x_{n+1}}{2} \le 1 \;\iff\; \frac{x_{n}}{2} \le 1 \;\iff\; \cdots \;\iff\; \frac{x_1}{2} \le 1\,$

• $\;\displaystyle \frac{x_{n+1}}{x_n}=\sqrt{\frac{x_{n}}{x_{n-1}}}\,$, so $\;\displaystyle \frac{x_{n+1}}{x_n} \ge 1 \;\iff\; \frac{x_{n}}{x_{n-1}} \ge 1 \;\iff\; \cdots \;\iff\; \frac{x_2}{x_1} \ge 1\,$