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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.

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  • $\begingroup$ 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
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    $\begingroup$ $\limsup$ 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
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    $\begingroup$ @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
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Even values of x are zero, odd values are one.

y is the opposite.

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    $\begingroup$ 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

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