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Let $x_1=a>0$ and $x_{n+1}=x_{n}+\frac{1}{x_{n}};n>1$. Then does the sequence $(x_n)$ converge?

The sequence $(x_n)$ is increasing. But I could not show that it is bounded. Any hint in this regard will be appreciated.

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Suppose it converge, and denote $\ell$ the limit. Clearly $\ell\neq 0$ (why ?)

Then, $$\lim_{n\to\infty }x_{n+1}=\lim_{n\to\infty }x_n+\lim_{n\to\infty }\frac{1}{x_n}\iff \ell=\ell+\frac{1}{\ell}\iff \frac{1}{\ell}=0\iff 1=0$$

contradiction !

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  • $\begingroup$ I also thought like this...but can I show that it is unbounded? $\endgroup$ – Anupam Jan 13 '15 at 14:14
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    $\begingroup$ if you proved that the sequence is increasing and that the sequence has no limit, you proved that the sequence is unbounded ! $\endgroup$ – idm Jan 13 '15 at 14:25
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Suppose $(x_n)$ is convergent $\iff$ Cauchy.
For each $n\in\mathbb{N}$ there exist a positive integer $K_n$ such that $$|x_p-x_q|\le\dfrac{1}{n}$$ whenever $p,q\ge K_n.$ For $p>q$ we have $$x_p-x_q=(x_p-x_{p-1})+(x_{p-1}-x_{P-2})+\cdots+(x_{q+1}-x_q)$$ $$x_p-x_q=\dfrac{1}{x_{p-1}}+\dfrac{1}{x_{p-2}}+\dots+\dfrac{1}{x_{q}}\le\dfrac{1}{n}$$ Hence $$\dfrac{1}{x_{r}}\le\dfrac{1}{n},\forall r\ge K_n.$$ This implies, for each $n\in\mathbb{N}$ there exist $K_n\in\mathbb{N}$ such that $x_r\ge n, \forall n\ge K_n.$
$(x_n)$ is unbounded. This is a contradiction.
Therefore $(x_n)$ is divergent.

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