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I know there is an answer to this question. But I have a different way to prove this. Can someone help me to check if my proof is correct? Thanks.

Proof: Suppose the sequence $x_1,x_2,x_3,...$ of real numbers given by $x_1=1$ and $x_{n+1} = x_n +\frac{1}{x_n^2}$ for each $n=1,2,3,...$ is bounded.

Then $$\lim_{n\to \infty} x_n=x$$ for some $x\in \mathbb{R}$.

Hence $$\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} (x_n +\frac{1}{x_n^2})=x+\frac{1}{x^2} \not = x=\lim_{n\to \infty} x_n.$$

So the supposition is false, and the sequence is unbounded.

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    $\begingroup$ Given that the sequence is bounded, you need to say one more thing before you can conclude it converges... $\endgroup$ – Wojowu Oct 8 '17 at 20:58
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    $\begingroup$ Pedagogically I recommend you to explicitly mention that $x \neq 0$. Other than that, your solution is fine. But I see no difference between your solution and the other solutions in the link. $\endgroup$ – Sangchul Lee Oct 8 '17 at 20:58
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    $\begingroup$ You should have $\frac{1}{x^2}$ not $\frac{1}{x}$ just left of $\neq$. $\endgroup$ – Rob Arthan Oct 8 '17 at 21:01
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$(x_n) $ is an increasing sequence.

if it is bounded above, it will converge and $x_{n+1}-x_n $ will go to zero.

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

cannot go to zero. thus $(x_n) $ is unbounded above.

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    $\begingroup$ Elegant argument! $\endgroup$ – Michael Lee Oct 8 '17 at 21:05

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