Let $(a_n)_n$ be a Cauchy sequence of rational numbers. Prove that $\dfrac{4a_n^4}{2 + a_n^3}$ is also a Cauchy sequence.

I know that Cauchy sequences can be added, subtracted and multiplied but I'm not sure how to apply this to the proof.


You can't. Let $(a_n)_n$ be a sequence of rational numbers converging to $-\sqrt[3]{2}$. (The choice of rational $a_n$ is possible thanks to the density of $\Bbb Q$ in $\Bbb R$.) Since it's convergent, it's Cauchy, and thus bounded.

The fraction $\dfrac{4a_n^4}{2+a_n^3}$ has bounded nominator, but its denominator tends to zero as $n$ tends to infinity. Therefore, this fraction is not bounded, thus it's not Cauchy.

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  • $\begingroup$ So to prove that it's Cauchy I would need to prove that it's bounded? so for example if the problem was $\dfrac{4a_n^4}{2 + a_n^2}$ then both the nominator and denominator would be bounded? $\endgroup$ – kat Mar 14 '18 at 16:58
  • $\begingroup$ @kat Boundedness is a necessary condition for a sequence to be Cauchy, but not a sufficient condition. Take $(-1)^n$ as an example. For your fraction in the comment, take $a_n \to \pm\sqrt2$ instead to see that the denominator tends to zero, so by the same reasoning, it's not Cauchy. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 14 '18 at 17:03
  • $\begingroup$ @GNUSupporter I don't think $a_n\to\pm\sqrt2$ causes any difficulty in the $4a_n^4/(2+a_n^2)$ example. $\endgroup$ – Angina Seng Mar 14 '18 at 17:14
  • $\begingroup$ @LordSharktheUnknown Thanks for comment. You're right. I confused $a_n^2 + 2$ with $a_n^2 - 2$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 14 '18 at 17:16
  • $\begingroup$ @kat Sorry for my confusion about the deominator. For $\dfrac{4a_n^4}{2 + a_n^2}$, yes, both the numerator and denominator are bounded. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 14 '18 at 17:22

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