Proof that any number is equal to $1$

Question

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?

Answer 1

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

Answer 2

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 $x=1$.

Answer 3

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.