# Example that $\lim \sup (x_n\cdot y_n)<\lim \sup (x_n)\cdot \lim \sup (y_n)$

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.

• 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. – vrugtehagel Mar 17 '16 at 14:36
• This is absolutely fine. – Alex M. Mar 17 '16 at 14:37
• $\limsup$ 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
• @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

## 1 Answer

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

y is the opposite.

• 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
• 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