Let $s_1=\sqrt{2}$ and let $s_{n+1}=\sqrt{2} \sqrt{s_n}$ for $n \ge 2.$ Then to prove: $$\lim_{n\rightarrow \infty }\{s_n\}=2$$
My attempt: it is found by induction that $s_n\le 2$ for $n \ge 2.$
And $$s_n = 2^{1/2 + 1/4 + ... +1/2^{n}}$$
$$s_{n+1}=2^{1/2 + 1/4 + ... +1/2^{n+1}}= s_n 2^{1/2^{n+1} } $$
From this how can we say $s_{n+1}\ge s_n?$
And how can we prove that this sequence converges to 2?
If a series is bounded and monoton increasing it converges. You find the limit by $lim s_n=lim s_{n+1}=s$