What I got right now is: $a_0 = \sqrt{1 + \sqrt{2}}$, $a_1 = \sqrt{1 + \sqrt{2 + a_0}}$, and $a_n = \sqrt{1 + \sqrt{2 + a_{n-1}}}$. I can say $a = \sqrt{1+\sqrt{2 + a}}$, right? If yes, how to show that $a = \sqrt{1+\sqrt{2 + a}}$?
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Unravel it by first squaring a, subtracting 1, squaring again, subtracting 2. Then you have a on the right and you can solve the polynomial.. – Eleven-Eleven Oct 30 '16 at 01:54
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you can say $a=\sqrt{1+\sqrt{2+a}}$ if the limit of $a_n$ exists, so you only need to show the limit exists somehow. Then take the limit as $n\to\infty$ on both sides to get the equality – PLE Oct 30 '16 at 01:54
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that is the part I don't know... how to show that limit exists... – Vincent Zheng Oct 30 '16 at 01:56
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1Would increasing and bounded above help? – Camille Oct 30 '16 at 01:58
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@VincentZheng compare it to sqrt(2+sqrt2+(... – Jacob Wakem Oct 30 '16 at 04:03
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@Alephnull What for? – Did Oct 30 '16 at 06:08
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@Did boundedness – Jacob Wakem Oct 30 '16 at 12:26
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@Alephnull Sorry but I do not get the argument. One will have, anyway, to show that "sqrt(2+sqrt(2+sqrt...", whatever that means, is bounded -- and the argument to do that is not simpler than to solve the case the OP asks about. – Did Oct 30 '16 at 12:53
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@Did it is wel known. – Jacob Wakem Oct 30 '16 at 12:56
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@Alephnull Yeah, but the value of this argument is zero since the exercise the OP asks is also well known... – Did Oct 30 '16 at 12:58
2 Answers
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The sequence is bounded above by $2$: certainly $a_0<2$ since $\sqrt{2}<3$
Once you know $a_{n-1}<2$, $a_n<\sqrt{1+\sqrt{2+2}}=\sqrt{3}<2$
increasing: $a_1>a_0$ clearly, and use induction again
So bdd above, incerasing -> convergent
PLE
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Consider for $x\geq 0$ the map $$ f(x)=\sqrt{1+\sqrt{2+x}}.$$ The derivative of the map is $$ 0\leq f'(x)=\frac{1}{4\sqrt{1+\sqrt{2+x}}\; \sqrt{2+x}}\leq \frac{1}{4}$$ By the MVT $|f(x)-f(y)|\leq \frac{1}{4}|x-y|$ is a Lipschitz contraction of ${\Bbb R}_+$ which is a complete metric space. Whence the sequence $a_{n+1}=f(a_n)$, $a_0=0$ converges to a unique fixed point of $f$.
H. H. Rugh
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