I want to answer the following question:
A real sequence defined by $x_{n+1}=x_{n}+x_{n}^{-1}, \ x_1 >0$ converges?
I think it's not true, but I can't find a counterexample. Can someone help me?
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why do you think it is not true? – Arnaldo Jul 11 '17 at 21:36
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Think about what is happening to the terms. Are they increasing or decreasing? Consider different starting numbers for $x_1$ – Jonathan Davidson Jul 11 '17 at 21:37
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I think it's not true because the terms are increasing. – Isabela Jul 11 '17 at 21:39
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@ivmat That doesn't necessarily imply divergence: consider the sequence $a_n=\frac{n}{n+1}$. – Carl Schildkraut Jul 11 '17 at 21:40
2 Answers
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In fact it isn't because if $\lim_{n\to \infty} x_n=L\in \Bbb R$ then
$$x_{n+1}=x_{n}+x_{n}^{-1}$$
would give you (when you make $n\to \infty$)
$$L=L+L^{-1}\to L^{-1}=0$$
what is impossible.
Arnaldo
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Hint:
$$x_{n+1}^2 = x_n^2+2+\frac{1}{x_n^2} > x_n^2+2.$$
Carl Schildkraut
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In fact, this helped me to bootstrap to convince myself $x_n \sim \sqrt{2n}$ as $n \to \infty$. – Daniel Schepler Jul 11 '17 at 22:03
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@DanielSchepler: https://math.stackexchange.com/questions/29777/closed-form-for-the-sequence-defined-by-a-0-1-and-a-n1-a-n-a-n-1?noredirect=1&lq=1 – Aryabhata Jul 12 '17 at 21:08