3
$\begingroup$

a monotonic sequence $\{x_n\}$ in $\mathbb{R}$ is convergent iff the sequence $\{(x_n)^2\}$ is convergent.

Forward Implications is easy, nothing to worry about that. I am confused about backward implication For backward Implication my attempt.

Attempt(Please check if this reasoning is True)

Case 1:- $\{x_n\}$ is Increasing. We have two possibilities further.

Either all $x_n<0$ or $x_n >0$ for n>N, where N is some natural number.

When all $x_n<0$ , we have it bounded above by 0. Hence its convergent.

Now if $x_n >0$ for n>N, where N is some natural number. Then because $\{(x_n)^2\}$ is convergent implies $(\{x_n)^2\}$ is bounded above. Hence $\{x_n\}$ is also bounded above. Hence $\{x_n\}$ is convergent.

Simmilar for when $\{x_n\}$ is decreasing.

If its not correct, Tell me the solution

$\endgroup$
  • 1
    $\begingroup$ This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it. $\endgroup$ – Clayton Jul 26 '18 at 23:20
  • 1
    $\begingroup$ This argument is perfect. $\endgroup$ – user578878 Jul 26 '18 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.