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$\lim_{n \to \infty}a_n=\sqrt{2+\sqrt{2+\sqrt{2}+\sqrt{2...}}}$

I am trying to find the limit using the Monotone convergence theorem

But first I am trying to find an explicit presentation of $a_n$.

is the only presentation of it is a sequence $$\sum_{n=1}^{\infty}2^\frac{1}{2n}$$ ?

gbox
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3 Answers3

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Hints:

$$a_n:=\overbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt2}}}^{n\;\text{times}}\implies \begin{cases}a_n\le a_{n+1}\\{}\\a_n\le 2\end{cases}\;\;\forall n$$

Prove then the limit exists and must be 2.

Timbuc
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  • using induction I have proved that it is Monotone, but how can I prove that $a_n\leq 2$ by induction if I have not find $a_n$ explicitly? – gbox Jan 13 '15 at 13:36
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    Induction @gbox: $$a_1=\sqrt2\le 2;,;;a_{n+1}=\sqrt{2+a_n}\le \sqrt{2+2}=;etc.$$ – Timbuc Jan 13 '15 at 13:37
  • = etc. means by the induction assumption $a_n\leq 2$ therefore $\forall_{n\in \mathbb{N}}a_n\leq 2$? – gbox Jan 13 '15 at 13:42
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    @gbox In fact, that "etc." means that $;\sqrt{2+2}=2;$ ... :) But yes, also that $;a_n\le 2;$ by the Ind. Assumption. – Timbuc Jan 13 '15 at 13:42
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$A=\sqrt{2+\sqrt{2+\sqrt{2}+\sqrt{2...}}}$

$A^2=2+A$

Yuko
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    The second equality is true if you can first prove the expression for $;A;$ exists (finitely). – Timbuc Jan 13 '15 at 13:25
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$a_{n+1} = \sqrt{2+a_n}$ and $a_0 = 2$. The limit $l$ must by continuity satisfy $l = \sqrt{l+2}$. Therefore $l^2 = l +2$. So the limit is the positive root of $X^2 - X - 2$. (Note that the two root have opposite sign, as their product is $-2$.) I leave you finish. ;-)

Olórin
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  • As Timbuc already said below, for that to be true, it needs to be shown that $a_n$ converges. – ryagami Jan 13 '15 at 13:28
  • @Robert Beside the fact that it must be proven (by you or, better, by the OP) that the limit exists, whose continuity are you referring to in your first line? – Timbuc Jan 13 '15 at 13:29
  • "I'm trying to find the limit" makes me assume that the OP knows the limit exists. And it's the continuity of the dimension of the fibers I'm referring to, as well as to the continuity of $x\mapsto\sin\left(\frac{1}{x}\right)$ on its definition domain. – Olórin Jan 13 '15 at 13:31
  • What? What fibers (of what) and what $;\sin\frac1x;$ are you talking about? Perhaps you did comment on the wrong question?! – Timbuc Jan 13 '15 at 13:33
  • No, I'm just nicely making fun of you. Because I guess that the OP got that it was continuity of $x\mapsto \sqrt{x+2}$ I was writing about. – Olórin Jan 13 '15 at 13:34
  • Oh. That was hilarious (?) – Timbuc Jan 13 '15 at 13:42
  • It really was. Especially because I'm sure you know that there's no continuity of the fibers, but only upper continuity. This is the semicontinuity theorem. Valid as you know of course for proper morphisms of noetherian schemes and for coherent and flat sheaves on the base. I'm laughing again. Thanks so mmuch ! – Olórin Jan 13 '15 at 13:48