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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, SchrodingersCat, Michael Albanese
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.