If $S_1 = \sqrt{2}$, and
$S_{n+1} = \sqrt{2 + \sqrt{S_n}}$ (n = 1,2,3....),
prove that $\{S_n\}$ converges, and that $S_n < 2$ for all $n \in \Bbb{N}$
This is one the questions from Principles of Mathematical Analysis by Rudin. I am not sure how to proceed with the solution.
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1Hint: After you establish boundedness ($S_n <2$ ), show monotonicity and deduce convergence. – user10444 Oct 14 '14 at 00:07
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possible duplicate of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ – Belgi Oct 14 '14 at 00:07
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Are you sure that it should be $S_{n+1}=\sqrt{2+\sqrt{S_n}}$? Could it be $S_{n+1}=\sqrt{2+S_n},$ perhaps? – Cameron Buie Oct 14 '14 at 00:09
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@Belgi: As currently written, it is not a duplicate. – Cameron Buie Oct 14 '14 at 00:10
2 Answers
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Hints:
1) $S_n < 2.$ Clearly, $S_1<2$. Suppose that $S_k <2$. Then $S_{k+1}=\sqrt{2 + \sqrt{S_k}} <\sqrt{2 + 2}=2.$
2) $S_n \le S_{n+1}$. Clearly, $S_1<S_2$. Suppose that $S_{k-1} <S_k$. Then $$S_{k}=\sqrt{2 + \sqrt{S_{k-1}}} <\sqrt{2 +\sqrt{S_{k}}}=S_{k+1}.$$
3) So its limit exists, say $S$. So, $S=\sqrt{2+\sqrt{S}}.$
Paul
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Well if we prove that $S_n<2$ for $n=1,2,3,\ldots$ then the first part comes easily as $S_n$ is a sequence in $\mathbb{R}$ which is complete. Since $S_n$ is a monotone increasing sequence with an upper bound (namely 2), it converges. We can prove $S_n<2$ for $n \ge 1$ via induction as follows:
$\textbf{I:}$ $ S_1 < 2 $, i.e. $\sqrt{2} < 2$, can be checked by direct computation.
$\textbf{II:}$ Suppose $S_{n-1}<2$ but that $S_n \ge 2$. But then we have $\sqrt{2 + \sqrt{S_{n-1}}} \ge 2 \Rightarrow \sqrt{S_{n-1}} \ge 2$, a contradiction. Thus $S_n < 2$.
$\textbf{III:}$ Thus $S_n<2$ for all $n = 1,2,3,\ldots$.
Kevin Sheng
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