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Let $x_1 > 0$ and

$ x_{n+1} = 1/2 (x_n + 2/x_n)= x^2n+2/2x_n, n\geq1.$

Does $\{x_{n+1}\}$ converge? If so, find its limit.

Hint: Prove first that if $a, b \geq 0$ ,then $ 2ab \leq a^2 + b^2. $

I don't see how the hint is suppose to help.

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Try formatting your expressions to avoid ambiguity, in particular your fractions. –  glebovg Nov 9 '12 at 4:33
    
Your equality does not make sense. Please edit. –  glebovg Nov 9 '12 at 4:37

1 Answer 1

up vote 0 down vote accepted

Suppose it converges, then both ${x_n}$ and ${x_{n+1}}$ approach the same limit, say $x$. This means (if I deciphered your expression correctly) $$x = \frac{1}{2}\left( {x + \frac{2}{x}} \right)$$ or $$2{x^2} = {x^2} + 2$$ so $$x = \pm \sqrt 2.$$ Since all terms are positive, the limit must be $\sqrt 2$.

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We can prove that $x_2\geq x_3\geq\ldots\geq\sqrt{2}$, so the sequence converges. –  Siming Tu Nov 9 '12 at 4:51

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