# Disproving that if $x^2_n$ is Cauchy, then $x_n$ is Cauchy

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.

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

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.
• 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? – wundering Dec 9 '19 at 2:52
• 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. – user284331 Dec 9 '19 at 2:54
• 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. – wundering Dec 9 '19 at 2:57
• 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. – user284331 Dec 9 '19 at 3:00