This is a short question, I already managed to prove using definitions that $$\lim \sup (x_n\cdot y_n)\le \lim \sup (x_n)\cdot \lim \sup (y_n)$$

But I'm having trouble coming up with an example such that $$\lim \sup (x_n\cdot y_n)<\lim \sup (x_n)\cdot \lim \sup (y_n)$$

I tried to consider alternative sequences but i'm not sure if i'm doing it right. I'm considering the following right now. $$x_n=(1,0,1,0,...)$$ $$y_n=(0,1,0,1,...)$$ $$x_n\cdot y_n=(0,0,0,0,...)$$ $\lim \sup x_n \cdot y_n=0$ as there sequence is convergent. But $\lim \sup x_n = 1$ and $\lim \sup y_n =1$ So it appears the inequality holds. I just need a confirmation that what i'm doing is right. Sorry if this is a redundant question, I'm just learning this concept so it's a little fuzzy for me.

Note that $(x_n)$ and $(y_n)$ are non-negative.

  • $\begingroup$ Exactly what are you unsure about? $\endgroup$ – A.S. Mar 17 '16 at 14:35
  • $\begingroup$ I don't see anything wrong with it. In fact, I think you understand the concept fairly well for someone who just started learning it. $\endgroup$ – vrugtehagel Mar 17 '16 at 14:36
  • $\begingroup$ This is absolutely fine. $\endgroup$ – Alex M. Mar 17 '16 at 14:37
  • 1
    $\begingroup$ $\limsup$ is the best "bound" at infinity. Just $\sup$ is not satisfactory - $x_0$ might be huge compared to the rest hence "unrepresentative". $\limsup$ gives you a "long-term/stable" best upper boundary - even though the sequence might always stay above it. It's a way to "squeeze" a sequence in some meaningful way. $\endgroup$ – A.S. Mar 17 '16 at 14:42
  • 1
    $\begingroup$ @Mambo: Your example fails: both sequences are convergent, therefore in your case $\limsup x_n y_n = \limsup x_n \limsup y_n$. $\endgroup$ – Alex M. Mar 17 '16 at 15:27

Even values of x are zero, odd values are one.

y is the opposite.

  • 1
    $\begingroup$ I got a -1 for a similar answer, which I deleted 14 minutes ago ... let's see if people dare to downvote this one too ;) $\endgroup$ – rtybase Mar 17 '16 at 15:54
  • $\begingroup$ Oh, come on, Marty, this is the OP's example! Take a look at the other deleted answers (you have >10k reputation, so you may see them) - they all present the same example. In fact, the OP wasn't even after it, he was asking for confirmation that his understanding (and example) is correct. $\endgroup$ – Alex M. Mar 17 '16 at 16:28
  • $\begingroup$ I think, the confusion arises from 2 factors: 1. Vector like notation $x_n=(1,0,1,0,...)$ 2. $\lim \sup (x_n)$ (also noted as $\overline{\lim} x_n$ in some countries) may be easily confused with $\lim ( \sup (x_n))$ (which is a limit of a constant assuming $\sup$ exists), e.g. @Mambo comment. So, let's not be too harsh ;) $\endgroup$ – rtybase Mar 17 '16 at 16:45

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.