Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 3

Problem

If $s_1 = \sqrt{2}$, and

$$ s_{n+1}=\sqrt{2 + \sqrt{s_n}}\;\;\;(n=1,2,3,\cdots),$$

prove that $\{s_n\}$ converges, and that $s_n < 2$ for $n=1,2,3,\cdots$.

Answer

Note that $s_1 < 2$. If $s_n < 2$ then

$$ s_{n+1} = \sqrt{2 + \sqrt{s_n}} < \sqrt{2+\sqrt{2}} < 2. $$

By mathematical induction, $s_n < 2$ for $n=1,2,3,\cdots$.

Now we claim that $s_n$ is increasing, and hence convergent.

Note that $s_1 = \sqrt{2} < \sqrt{2 + \sqrt{2}}$. If $s_{n-1} < s_n$ then

$$ s_{n+1} = \sqrt{2+\sqrt{s_n}} > \sqrt{2 + \sqrt{s_{n-1}}} = s_n$$.

By mathematical induction, $s_n$ is increasing.

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