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Let $x_1 = a > 0$, and let $x_{n+1} = x_n + 1/x_n$ for all $n\in\mathbb{N}$. Does $x_n$ converge or diverge?

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closed as off-topic by Andrés E. Caicedo, Zev Chonoles, AlexR, Hagen von Eitzen, user127.0.0.1 Feb 23 '14 at 16:53

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Can you show it's bounded? – Marcin Łoś Feb 23 '14 at 16:11
Have you at least looked at examples yourself? – KCd Feb 23 '14 at 16:12
@MarcinŁoś It's not. – Julien Godawatta Feb 23 '14 at 16:12
@JulienGodawatta I know. Perhaps that was a bit too indirect of a hint. – Marcin Łoś Feb 23 '14 at 16:16

If a sequence $x_n$ converges then the sequence $x_n^2$ should also converge.

Here $x_{n+1}^2=x_n^2+\frac{1}{x_n^2}+2$

Note that $|x_{n+1}^2-x_n^2|>2\Rightarrow x_n^2 $ diverges so does $x_n$.

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Ratio test for convergence of series (see Wikipedia). $$\lim_{n\rightarrow\infty}\left|\frac{x_{n+1}}{x_n}\right|=\lim_{n\rightarrow\infty}\frac{x_n+\frac{1}{x_n}}{x_n}=\lim_{n\rightarrow\infty}1+\frac{1}{x_n^2}$$ So either $x_n^2$ goes to $\infty$ or if it is not then $$\lim_{n\rightarrow\infty}1+\frac{1}{x_n^2}>1$$ But in that case $r>1$ and the sequence is divergent. So in both cases (either by assumption or by the ration test) you have that the sequence is divergent (that is $x_n \rightarrow \infty$ as $n \rightarrow \infty$).

(Perhaps you can show additionally tha $x_n^2>0$ but that follows (by induction) from the fact that $x_1=\alpha>0$.)

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Now that I see it again, in the first case I have that $x_n^2$ is divergent not that $x_n$ is divergent. So it can be that $x_n^2$ diverges but $x_n$ not, so in that case the ratio test is inconclusive. – Jimmy R. Feb 23 '14 at 16:36
if the sequence $x_n^2$ diverges, then the sequence $x_n$ must also diverge. Clearly, if $x_n^2$ is unbounded, then so is $x_n$. – Omnomnomnom Feb 23 '14 at 16:38
@Omnomnomnom Yes, ok now I have it. If $x_n$ is bounded by $M$ then $x_n^2$ can not be unbounded since it is bounded by $M^2$. – Jimmy R. Feb 23 '14 at 16:47

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