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If $a_n$ is a Cauchy sequence, show $\dfrac{3a_n^3}{1+a_n^2}$ is also Cauchy.

So far I have:

$$\left|\frac{3a_m^3}{1+a_m^2} - \frac{3a_n^3}{1+a_n^2}\right| = \left|\frac{3a_m^3 + 3a_m^3a_n^2-3a_n^3-3a_n^3a_m^2}{(1+a_m^2)(1+a_n^2)}\right|$$

I'm not really sure how to proceed from here.

Thanks in advance for the help.

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  • $\begingroup$ Why is this downvoted? $\endgroup$
    – TheSimpliFire ♦
    Commented Mar 14, 2018 at 7:58
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    $\begingroup$ Have you learned yet that in $\mathbb {R}$, Cauchy Sequences are convergent and vice versa? Because if you have, you can go around using an $\epsilon$ argument. $\endgroup$
    – Ovi
    Commented Mar 14, 2018 at 8:22

2 Answers 2

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Hint. Let $f(x)=3x^3/(1+x^2)$ then $$f(x)-f(y)=3\cdot \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\cdot (x-y).$$ Show that $$0\leq \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\leq 2,$$ that is $$0\leq x^2y^2+x^2+xy+y^2\leq 2x^2y^2+2x^2+2y^2+2.$$ Hence $|f(a_n)-f(a_m)|\leq 6 |a_n-a_m|$.

P.S. More generally if $(a_n)_n$ is a Cauchy sequence then it is bounded: there is $M>0$ such that $|a_n|\leq M$ for all $n\in\mathbb{N}$. Moreover if $f$ is a $C^1(\mathbb{R})$ function then $f'$ is bounded over the compact set $[-M,M]$. Hence, by the Mean Value Theorem, if $a_n\not= a_m$, then there exists $t\in [-M,M]$ (which depends on $a_n$ and $a_m$) such that $f(a_n)-f(a_m)=f'(t)(a_n-a_m).$ Hence $$|f(a_n)-f(a_m)|\leq C |a_n-a_m|$$ where $C:=\max_{x\in [-M,M]}|f'(x)|$.

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  • $\begingroup$ Thank you for the answer. Unfortunately, I haven't learned how to apply the Mean Value Theorem. Is there a way to solve this by using N and epsilon? $\endgroup$
    – user540611
    Commented Mar 14, 2018 at 8:21
  • $\begingroup$ @user540611 See my P.S. $\endgroup$
    – Robert Z
    Commented Mar 14, 2018 at 8:33
  • $\begingroup$ Hmm how do you prove that big inequality? $\endgroup$
    – Ovi
    Commented Mar 14, 2018 at 8:44
  • $\begingroup$ @Ovi It is equivalent to $0\leq x^2y^2+(x^2-xy+y^2)+2$ (recall that $xy\leq x^2+y^2$ $\endgroup$
    – Robert Z
    Commented Mar 14, 2018 at 8:47
  • $\begingroup$ Ah right, I was trying to prove it directly. If you don't mind me asking: when you solved the problem did you just assume that rational function is bounded, then set $\frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\leq M$, and then "solved for $M$" by multiplying both sides by the denominator? $\endgroup$
    – Ovi
    Commented Mar 14, 2018 at 8:50
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If $\{a_n\}$ is Cauchy, then it converges. Say $a_n\to a\in\mathbb R$. Then clearly, $$ \frac{3a_n^3}{a_n^2+1}\to\frac{3a^3}{a^2+1} $$ and thus, since $\,\big\{\frac{3a_n^3}{a_n^2+1}\big\},\,$ is a convergent sequence, it is a Cauchy sequence, as well.

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