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I tried to work backwards from the fact that $x^2_n$ is cauchy to find a case where $x_n$ is not cauchy.

$\vert x^2_n-x^2_m \vert \lt \epsilon$

$\vert (x_n-x_m)(x_n+x_m) \vert \lt \epsilon$

$\vert x_n-x_m \vert \vert x_n+x_m \vert \geq \vert (x_n-x_m)(x_n+x_m) \vert \lt \epsilon$ (proved earlier in class)

Then a case would exist that $\vert x_n-x_m \vert \gt \epsilon/(\vert x_n+x_m\vert)$, making $x_n$ not a Cauchy sequence since $\epsilon/(\vert x_n+x_m \vert)\gt0$

Is this a legitimate approach, or are there assumptions that aren't allowed? Just looking for feedback as this is for a graded assignment that I can only seek guidance, not solutions for.

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    $\begingroup$ Are you assuming $x_n \ge 0$ for all $n$? $\endgroup$ – Henry Dec 9 '19 at 2:46
  • $\begingroup$ @Henry I am not, I reasoned that because of the absolute value it would be irrelevant if $x_n \lt 0$ $\endgroup$ – wundering Dec 9 '19 at 2:54
  • $\begingroup$ Your third line looks strange to me. If $|(x_n-x_m)(x_n+x_m)|\lt \varepsilon$, then $|x_n-x_m| \lt \dfrac{\varepsilon}{|x_n+x_m|}$ $\endgroup$ – Divide1918 Dec 9 '19 at 4:13
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You cannot have variable pairing with $\epsilon$.

A counterexample is the sequence defined by $(1,-1,1,-1,...)$, the absolute value is the constant sequence.

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  • $\begingroup$ I see. So would my solution need to somehow deduce that some $\epsilon$ not dependent on a variable is less than $\vert x_n-x_m \vert$ or should I not be starting with the fact that $x^2_n$ is Cauchy? $\endgroup$ – wundering Dec 9 '19 at 2:52
  • $\begingroup$ The philosophy of $\epsilon$-argument is that, the final step of $\epsilon$ inequality cannot end up with variable pairing with it, this is very crucial. $\endgroup$ – user284331 Dec 9 '19 at 2:54
  • $\begingroup$ Would it be the same process as proving that a sequence is Cauchy, just instead of $ \lt \epsilon$, I find that it is $ \gt \epsilon$? I'm not familiar with showing that sequences aren't Cauchy as I've almost exclusively dealt with showing that sequences are Cauchy. $\endgroup$ – wundering Dec 9 '19 at 2:57
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    $\begingroup$ If you end up something with $>\epsilon$, then it is not Cauchy, but usually this is not a very easy way, usually we select a concrete $\epsilon$ as a number and show that $>(\text{a number})$ and then conclude that it is not Cauchy. $\endgroup$ – user284331 Dec 9 '19 at 3:00

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