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It is clear that $\sup\{\sqrt2,\sqrt{2+\sqrt2},\cdots\}=2$. However, I am struggling to prove this without using limits or sequences. I know that I need to show that the set bounded above, and then we can determine that it has a supremum which I can then show is equal to $2$. How can I show that this set is bounded above in an elementary way? We have not yet covered sequences, limits, or convergence in my analysis class, nor have we covered the trigonometric functions.

csch2
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    induction: $\sqrt{2+\sqrt{2+\sqrt{2+\dots}}} \le 2$ iff $2+\sqrt{2+\sqrt{2+\dots}} \le 4$ iff $\sqrt{2+\sqrt{2+\dots}} \le 2$, where there are $n+1$ $2$'s appearing in the LHS at first expression and $n$ $2$'s appearing in the LHS in the last expression. – mathworker21 Aug 29 '19 at 03:23
  • Alternatively you can prove by induction that the term with $k$ nested square roots is equal to $2\cos(\pi/2^{k+1})\le2$. All you need is the half-angle formula. But that is largely coincidental and applies to this particular sequence only. – Jyrki Lahtonen Aug 29 '19 at 03:33
  • @mathworker21 That is a standard (but rather nice) way of proving that $2$ is an upper bound. Proving that it is the least upper bound is a bit trickier without the theorem of convergence on bounded monotonous sequences. – Jyrki Lahtonen Aug 29 '19 at 03:52
  • The sequence has been studied in many a thread already. Start for example here, and check out all the other threads linked to it et cetera. Among them you will find a proof for the formula I gave. – Jyrki Lahtonen Aug 29 '19 at 04:00
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    Possible duplicate of Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$ – Jyrki Lahtonen Aug 29 '19 at 04:01
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    @JyrkiLahtonen OP just asked for pf of upper bound, which I provided. Also, how is OP's question a duplicate of what you linked to? Of course what you linked to implies an upper bound, but the link and the OP's question are by no means equivalent. – mathworker21 Aug 29 '19 at 04:06
  • @mathworker21 1) I did upvote your comment. 2) The title question asks for the supremum. 3) The other threads studying this sequence are MUCH more informative about sequences, but they all assume the theorem on monotone sequences. My target of choice implies the supremum immediately. You are welcome to disagree. – Jyrki Lahtonen Aug 29 '19 at 04:10
  • @JyrkiLahtonen I have edited the question title to more clearly reflect what I am asking in the question body. Also, thank you for the other linked question, but unfortunately we have not yet covered/constructed the trigonometric functions. – csch2 Aug 29 '19 at 04:18
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    @JyrkiLahtonen I agree with what you have said; I was just objecting to "duplicate". – mathworker21 Aug 29 '19 at 04:20

2 Answers2

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How can I show that this set is bounded above in an elementary way?

Denote by $a_k$ the general term: $a_1 = \sqrt{2}$, $a_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$ etc.

Suppose $a_k \leq 2$. Then $$a_{k+1} = \sqrt{2+a_k} \leq \sqrt{2+2} = 2 $$ I don't know of any elementary (read: no sequences or convergence theorems) way of showing it is the least upper bound.

AlvinL
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You can prove as follows: $$a_n=\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt{2+\sqrt{2+\color{red}{\sqrt{2}}}}}}}}_{n} <\\ \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt{2+\sqrt{2+\color{red}{2}}}}}}}_{n}=2$$

farruhota
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