From Fitzpatrick's Advanced Calculus book: "Suppose that the sequence ${\{a_n}\}$ is monotone, i.e., either monotonically increasing or decreasing. Prove that ${\{a_n}\}$ converges if and only if ${\{{a_n}^2}\}$ converges. Show that this result does not hold without the monotonicity assumption."

It is easy to show to that ${\{{a_n}^2}\}$ converges if ${\{a_n}\}$ converges: ${\{a_n}\}$ is convergent sequence and ${\{a_n}\}$ also, since product of two convergent sequences converges (to the product of their limit), so ${\{{a_n}^2}\}={\{a_n\times a_n}\}$ converges; and it doesn't require that ${\{a_n}\}$ to be monotone.

On the other hand, it is not general true to say that ${\{a_n}\}$ converges if ${\{{a_n}}^2\}$ converges; ${\{a_n}\}={\{(-1)^n}\}$ is a counterexample.

What remains to evaluate (i.e., my questions) is that:

$1-$ Prove that if ${\{{a_n}^2}\}$ converges and if ${\{a_n}\}$ is monotone, thus ${\{a_n}\}$ converges.


$2-$ If ${\{{a_n}^2}\}$ converges to $a^2$, to which of $a$ or $-a$ does ${\{a_n}\}$ converge?

Thank you.

  1. If $\{a_n^2\}$ converges, it is bounded. Then $\{a_n\}$ is bounded and monotone.

  2. Anything can happen. There is no way to know a priori.

  • $\begingroup$ Yes, The Monotone Convergence Theorem ! I can see that both of $a$ and $-a$ are possible; is there a way to prove that the limit of ${\{a_n}\}$ is only a member of ${\{a, -a}\}$? $\endgroup$ – L.G. Jan 28 '15 at 13:21
  • 1
    $\begingroup$ If $a_n\to\ell$ then $a_n^2\to\ell^2=a^2$. $\endgroup$ – Julián Aguirre Jan 28 '15 at 13:22
  • $\begingroup$ How do you show if ${\{a_n^2\}}$ is bounded then ${\{a_n\}}$ is bounded also? $\endgroup$ – Sun Mar 30 at 4:02
  • $\begingroup$ @Sunny $|a_n|^2\le M\implies |a_n|\le\sqrt M$. $\endgroup$ – Julián Aguirre Mar 30 at 6:01

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