0
$\begingroup$

Given the sequence of real numbers $\{x_n\}_{n \in \mathbb N}$, we define $\{y_n\}_{n \in \mathbb N}$ where $y_n=\max\{|x_1|,...,|x_n|\}$ for each $n \in \mathbb N$. Prove that if $\{x_n\}_{n \in \mathbb N}$ is bounded, then $\{y_n\}_{n \in \mathbb N}$ is a convergent sequence.

My attempt at a solution.

If $\{x_n\}_{n \in \mathbb N}$ is bounded, then, it has a convergent subsequence. Call that sequence $\{x_{n_k}\}_{k \in \mathbb N}$. Note that if $x=\lim_{k \to \infty} x_{n_k}$, then $|x|=\lim_{k \to \infty}|x_{n_k}|$. This can be proved by the fact that $0\leq ||x_{n_k}|-|x||\leq |x_{n_k}-x| \to 0$ when $k \to \infty$.

I was going to try to prove that $|x|=lim_{n \to \infty} y_n$ but immediately realize that this doesn't need to be true. For example: $\{x_n\}_{n \in \mathbb N}$: $x_n=0$ if $n$ is odd and $x_n=1$ if $n$ is even has two convergent subsequences.

My problem is I don't know what else to do, I would appreciate some guidance.

$\endgroup$
  • 4
    $\begingroup$ To show that $y_n$ is convergent, focus on the properties of $y_n$. There are several facts about $y_n$ that you should be able to deduce directly from your above statements. Do those facts help you to prove convergence? $\endgroup$ – John Dec 20 '13 at 19:40
  • $\begingroup$ @John right, $\{y_n\}_{n \in \mathbb N}$ is bounded and it is monotone increasing, I can't believe I didn't realize it before $\endgroup$ – user100106 Dec 20 '13 at 19:43
  • $\begingroup$ @user100106 : the MSE system dislikes unanswered questions. You can answer your own question, or John can answer it. $\endgroup$ – Stefan Smith Dec 20 '13 at 20:04
1
$\begingroup$

To show that $y_n$ is convergent, focus on the properties of $y_n$. There are several facts about $y_n$ that you should be able to deduce directly from your above statements. Do those facts help you to prove convergence?

$\endgroup$
2
$\begingroup$

It is consequence of the following:

If a sequence is monotonic and bounded, then it converges.

$\endgroup$
  • $\begingroup$ As realized by the OP 50 minutes before this was posted. $\endgroup$ – Did Dec 20 '13 at 21:42

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.