Exercise 3.3

Question

Exercise 3:
    If $s_1=\sqrt{2}$, and $$s_{n+1}=\sqrt{2+\sqrt{s_n}}\quad\quad(n=1,2,3,\ldots),$$ prove that $\{s_n\}$ converges, and that $s_n<2$ for $n=1,2,3,\ldots$.

ghostofgarborg · Accepted Answer

Note that if $s_n < 2$, $s_{n+1} = \sqrt{2 + \sqrt{s_n}} < \sqrt{4} = 2 $, so the sequence is bounded. Also note that if $s_{n-1} > s_{n-2}$, then $s_n = \sqrt{2 + \sqrt{s_{n-1}}} > \sqrt{2 + \sqrt{s_{n-2}}} = s_{n-1}$, so the sequence is monotonically increasing. It therefore converges by thm 3.14.

Exercise 3.3

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Exercise 3.3

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