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Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working: Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$ $$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-2=0\implies x=2$$ as $x>0$. Now for the proof: I was attempting some different infinite expansions/square roots when trying to solve another question of mine (Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ ) and I came across this: $$x+\frac{1}{x}=\sqrt{(x+\frac{1}{x})^2}=\sqrt{2+x^2+\frac{1}{x^2}}=\sqrt{2+\sqrt{(x^2+\frac{1}{x^2}}})^2=\sqrt{2+\sqrt{2+x^4+\frac{1}{x^4}}}=\sqrt{2+\sqrt{2+\sqrt{(x^4+\frac{1}{x^4})^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^8+\frac{1}{x^8}}}}=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=2$$ if you keep on applying this and using the result found at the start of the question. So we have that for any real number $x$ that $$x+\frac{1}{x}=2\implies x^2-2x+1=0\implies (x-1)^2=0$$ so we finally have: $$x=1$$ Where have I gone wrong, for surely this cannot be correct?

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    $\begingroup$ You are indirectly assuming that the square root sequence you build is converging (I.e. That you can at some point drop the tail with the x^(2n) which is not at all clear to me, I could also argue that x=1/2 +(x-1/2)=1/2+1/4+(x-1/2-1/4)=...=1/2+1/4...=1. $\endgroup$
    – Felix
    Commented Jul 8, 2020 at 20:35
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    $\begingroup$ I'll haunt high school children with this lol. $\endgroup$ Commented Jul 8, 2020 at 20:52
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    $\begingroup$ The mistake is that the $x$ defined in the fisrt calculus dont is the same $x$ that you used in the second radical. $\endgroup$ Commented Jul 9, 2020 at 0:52

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As you go along the last square root has $x^{2n}+\frac{1}{x^{2n}}$ which diverges, so it can't be ignored as $n\to \infty$

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  • $\begingroup$ $x^{2n}$ should be $x^{2^n}$. Doesn't effect answer.. $\endgroup$ Commented Jul 8, 2020 at 21:25
  • $\begingroup$ Surely that is like saying that the infinite square root $\sqrt{{x^2}+\sqrt{{x^4}+\sqrt{{x^8}+\sqrt{{x^{16}}...}}}}$ doesn't converge because the general term is $x^{2^n}$ which also diverges? However, the infinite square root I've mentioned definitely does converge; it's equal to $x(\frac{1+\sqrt5}{2})$ which is rather a lovely result incidentally. $\endgroup$ Commented Jul 8, 2020 at 22:10
  • $\begingroup$ @A-level Student You are right about the sequence you describe, but I suspect the extra term $\frac{1}{x^{2^n}}$ makes a big difference. $\endgroup$ Commented Jul 9, 2020 at 0:45
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Although they look similar at first glance, there is no reason for the two sequences $$ \sqrt{2+2},\ \sqrt{2+\sqrt{2+2}},\ \sqrt{2+\sqrt{2+\sqrt{2+2}}}, \ldots $$ and $$\sqrt{2+x^2+1/x^2},\ \sqrt{2+\sqrt{2+x^4+1/x^4}},\ \sqrt{2+\sqrt{2+\sqrt{2+x^8+1/x^8}}}, \ldots $$ to have the same limit unless $x=1$.

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  • $\begingroup$ Hi there, thanks for the answer. Would you mind explaining why they don't have the same limit though? $\endgroup$ Commented Jul 8, 2020 at 22:12
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    $\begingroup$ You proved that they don't. One is $2$, the other is $x+1/x$. But in general, when somebody claims two things are equal the onus is on them to prove they are equal, not on me to show why they aren't equal. $\endgroup$ Commented Jul 8, 2020 at 22:32
  • $\begingroup$ My apologies; that is what is my question is, for you to help me in proving that they are unequal; I have supplied my 'proof' of why they are equal and I am asking for help to disprove it. $\endgroup$ Commented Jul 8, 2020 at 22:34
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As other answers explained clearly, you cannot just replace repeated operation with ... and wave away the terms on the tail end without justification. If you still struggle to understand why this is wrong, what you did is roughly equivalent to:

$$ \begin{aligned} x + \frac 1 x &= 2 + (x + \frac 1 x - 2) \\ &= 2 + 0 + (x + \frac 1 x - 2) \\ &= 2 + 0 + 0 + (x + \frac 1 x - 2) \\ &= 2 + 0 + 0 + 0 + (x + \frac 1 x - 2) \\ &= 2 + 0 + 0 + 0 + ... \\ &= 2 \end{aligned} $$

Every partial sum equals to $x + \frac 1 x$ and we don't get to claim that the series converges to $2$ simply because we can insert an arbitrary number of repeated operation (here $+0$) in the middle.

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