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Prove that the sequence given by $x_1=1, x_{n+1}=x_n+\frac{1}{x_n^2}$ is unbounded.


It is enough to prove that $\lim_{n\rightarrow\infty} x_n = \infty$. Any hint please?

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closed as off-topic by mercio, Najib Idrissi, SchrodingersCat, luka5z, Michael Albanese Nov 30 '15 at 15:31

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    $\begingroup$ Possible duplicate of Proving $a_{9000}>30$ when $a_1=1$, $a_{n+1}=a_n+ \frac{1} {a_n^2}$ $\endgroup$ – mercio Nov 29 '15 at 10:36
  • $\begingroup$ How is this a duplicate? $\endgroup$ – luka5z Nov 29 '15 at 10:37
  • $\begingroup$ @luka5z Although proving unboundedness does not imply the result in the other question, but the answers contain ideas about how to constraint the growth rate to show that the sequence is increasing without bound. $\endgroup$ – Element118 Nov 30 '15 at 12:54
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Suppose $k<x_n\leq k+1$. We want to show that there exist $x_m>k+1$, so we can plug $m$ back in to find a $x_{m'}>k+2$ and so on.

Consider $x_{n+(k+1)^2}$. Suppose for sake of contradiction it is $\leq k+1$. Then using the fact that $\{x\}_{i=1}^\infty$ is strictly increasing, it follows that for all $n\leq l \leq n+(k+1)^2$, $k<x_l\leq k+1$. As such, $\frac{1}{k^2}>\frac{1}{x_l^2}\geq\frac{1}{(k+1)^2}$.

$x_{n+(k+1)^2}=x_n+\frac{1}{x_n^2}+\frac{1}{x_{n+1}^2}+\dots+\frac{1}{x_{n+(k+1)^2-1}^2}\geq x_n+\frac{1}{(k+1)^2}\times(k+1)^2=x_n+1>k+1$, a contradiction.

Hence, $x_{n+(k+1)^2}>k+1$, so $m=n+(k+1)^2$ suffice.

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