How can I prove that the supremum for the following set $S$ is equal to $2$ ?
$$S=\left\{\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\dots\right\}$$
My way is to let $a_1$ = $\sqrt{2}$ and $a_2$= $\sqrt{2 \sqrt{2}}$. Then I will rewrite $a_n+1$ = $\sqrt{2(a_n)}$ , for all $n\in \mathbb{N}$. Assume $a_k < 2 $, then $a_k+1 = \sqrt{2(a_k)} < \sqrt{2(2)} = 2$. Since $a_1 = \sqrt{2} < 2$, then by mathematical induction , $a_n < 2$. Hence $2$ is a supremum of $S$.
I not sure whether I do this right or not . Your help is very useful for me as I am going to have exam after this !
For any $\alpha\in(\sqrt{2},2)$ we have $\alpha<\sqrt{2\alpha}<2$, hence the given sequence is increasing and bounded, so convergent to some limit $L\in[\sqrt{2},2]$. Such limit has to fulfill $L=\sqrt{2L}$, hence $L=2$.