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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.

Eric Wofsey
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Danxe
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  • Exactly what are you unsure about? – A.S. Mar 17 '16 at 14:35
  • 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. –  Mar 17 '16 at 14:36
  • This is absolutely fine. – Alex M. Mar 17 '16 at 14:37
  • Well I'm really confused with this idea of lim sup and lim inf. Why do we need such definitions. I mean I understand the notion of supremum and infimum relates to the construction of real numbers but this one remains a mystery to me. – Danxe Mar 17 '16 at 14:37
  • Yes you are right. Look at the following sequences $x_n = 1+ \frac{1}{n}$ and $y_n = 1- \frac{1}{n}$. – Mambo Mar 17 '16 at 14:37
  • Thanks for your responses! – Danxe Mar 17 '16 at 14:40
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    $\limsup$ 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. – A.S. Mar 17 '16 at 14:42
  • @A.S. Hmm I think I can somewhat understand that. Thanks for taking the time to explain it man! – Danxe Mar 17 '16 at 14:48
  • The most apparent use is that to show a sequence converges, you need to show that $\limsup=\liminf$. – A.S. Mar 17 '16 at 14:49
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    @Mambo: Your example fails: both sequences are convergent, therefore in your case $\limsup x_n y_n = \limsup x_n \limsup y_n$. – Alex M. Mar 17 '16 at 15:27
  • @AlexM. Yes both are convergent to $1$. The example fails. My bad. – Mambo Mar 17 '16 at 16:02
  • @Mambo None the less, it is very interesting. I didn't consider such a sequence. Thanks for the input! – Danxe Mar 17 '16 at 16:05
  • Here is a post related specifically to the inequality at the beginning: lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n$. (As a side note, the condition that the sequences are non-negative is easy to miss, since it is only mentioned at the end of the post.) – Martin Sleziak Sep 25 '19 at 05:20

1 Answers1

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

y is the opposite.

marty cohen
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    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 ;) – rtybase Mar 17 '16 at 15:54
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    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. – Alex M. Mar 17 '16 at 16:28
  • 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 ;)

     – rtybase Mar 17 '16 at 16:45