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Find the value of $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$

I tried getting rid of square roots and got $(...((9-x)^2-x^2)^2-...)^2-x^n = 0$ which I don't think helped. Please point me in the right direction.

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    $\begingroup$ Did you mean for the limit to be as $n \to \infty$ instead of $x \to \infty$? $\endgroup$ – JimmyK4542 Feb 17 '16 at 7:40
  • $\begingroup$ I'm sorry, yes, I meant that. Edited $\endgroup$ – waterfalls Feb 17 '16 at 7:43
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    $\begingroup$ Answer is 4. Not sure how to prove it though. $\endgroup$ – Ian Miller Feb 17 '16 at 7:58
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    $\begingroup$ As Lucian pointed out, from this post we have the more general equality,$$\sqrt{y^2+\sqrt{4y^2+\sqrt{4^2y^2+\sqrt{4^3y^2+\dots}}}}=y+1$$ Let $y=1$, $$\sqrt{1+\sqrt{4+\sqrt{4^2+\sqrt{4^3+\dots}}}}=2$$ So the answer to your question is $x=4$. $\endgroup$ – Tito Piezas III Feb 17 '16 at 16:07
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Let me describe a sketch of proof that $x=4$.

A. Observe that if $f(x)=\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n}}}}$, then $f$ is strictly increasing.

B. We shall show that $f(4)=2$, and hence $x=4$ is the unique answer.

$B_1.$ Fix $m\in\mathbb N$ and show that, for $n=m,m-1,m-2,\cdots$ (induction backwards) $$ 2^n<\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}<2^n+1, $$ while $$ \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}=2^n+1. $$

$B_2.$ Next estimate the difference $$ (2^n+1)- \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\ =\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}- \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\ =\frac{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}- \sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}{\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}+ \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}} \\ <\frac{{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}- \sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}}{2\cdot 2^n} \\ <\cdots<\frac{(\sqrt{4^m}+1)-\sqrt{4^m}}{2^{m-n}\cdots 2^{n+(n+1)+\cdots+(m-1)}}=2^{-\frac{(m-n)(n+m+1)}{2}} $$ Thus $$ \lim_{m\to\infty}\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^n+1. $$ For $n=0$ we have $$ \lim_{m\to\infty}\sqrt{1+\sqrt{4+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^0+1=2. $$

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  • $\begingroup$ I have corrected it now! $\endgroup$ – Yiorgos S. Smyrlis Feb 17 '16 at 16:30
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Hint: Pretend that there's an extra $+1$ at the next-to-last level. Then $$4^{n-1}+\sqrt{4^n}+1~=~2^{2(n-1)}+2^n+1~=~2^{2(n-1)}+2\cdot2^{n-1}+1~=~(2^{n-1}+1)^2.$$ Can you see what happens ? :-$)~$ Now, as $n\to\infty,$ the numerical influence gained by adding that extra $+1$ at the top level tends towards $0.$

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$$A = \sqrt{1+\sqrt{x+\sqrt {x^2+\sqrt{x^3+\sqrt{x^4+\sqrt{x^5+...}}}}}} = 2$$

$$A = \sqrt{1+\sqrt{x\left(1+\sqrt {1+\sqrt{x^1+\sqrt{x^2+\sqrt{x^3+...}}}}\right)}} = 2$$

$$A = \sqrt{1+\sqrt{x(1+A)}} = 2,A =2$$

$$ \sqrt{1+\sqrt{3x}} = 2$$

$$ 1+\sqrt{3x} = 4$$

$$ \sqrt{3x} = 3$$

$$ x = 3$$

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    $\begingroup$ I am afraid that the second line is incorrect. $\endgroup$ – Lucian Feb 17 '16 at 9:24
  • $\begingroup$ The sequence should be $\sqrt{1+\sqrt{x}}, \sqrt{1+\sqrt{2x}}, \sqrt{1+\sqrt{x+\sqrt{x^2+x\sqrt{x}}}}, \sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{x^{3}+x^{2}}}}},\ldots$ and the above method doesn't work. $\endgroup$ – Ng Chung Tak Feb 17 '16 at 9:32
  • $\begingroup$ no, because A is a limit that has a finite response. $\endgroup$ – kevin Feb 17 '16 at 9:32
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    $\begingroup$ Check with the numerical values:$x=3$ wolframalpha.com/input/… and $x=4$ wolframalpha.com/input/… $\endgroup$ – Ng Chung Tak Feb 17 '16 at 9:35
  • $\begingroup$ I have also tried with $x=3.5$ link and $x=4.001$ link,you are correct answer is $x=4$. $\endgroup$ – john boehner Feb 17 '16 at 11:22

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