How to prove that $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }=2$
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Let $\sqrt{2+\sqrt{2+...}}=a$. Then squaring both sides we have $a^2=2+a$. This is a quadratic equation in $a$. Find the solutions of $a$. Then $a=2 $ or $a=-1$. Since $a>0, a=2$.
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1The second solution to the equation $a^2=2+a$ is not $-2$ but $-1$. – HorizonsMaths Sep 18 '15 at 05:30
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2You also have to show it converges. – marty cohen Sep 18 '15 at 13:55
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Exactly as @martycohen said: You've proved that if it has a solution, then 2 is the solution; but you need to prove the antecedent: how do you know it has a solution? – Matt Gutting Sep 18 '15 at 18:25
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Hint: if $\sqrt{2} < x < 2$, show that $\sqrt{2} < \sqrt{2+x} < 2$ and that $2-\sqrt{2+x} < 2-x$. – marty cohen Sep 18 '15 at 18:35