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I'm having problems with the following proof: If $s_n$ and $t_n$ are sequences, then does $\liminf (s_n\cdot t_n) = \liminf (s_n) \cdot \liminf (t_n)$?

Is there a theorem that proves this? Is this even true? If not, is there a counter example I could use to show that it is not true?

Martin Sleziak
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user115964
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    Let $s_n=-1$ if $n$ is odd and $s_n=0$ if $n$ is even, and let $t_n=0$ is $n$ is odd and $t_n=-1$ if $n$ is even. – Andrés E. Caicedo Dec 15 '13 at 17:09
  • can you explain what you mean by this? Is this a counter example or are you trying to prove the theorem? – user115964 Dec 15 '13 at 17:15
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    Think about it. – Andrés E. Caicedo Dec 15 '13 at 17:16
  • Likewise: $s_{2n}=t_{2n+1}=1$, $s_{2n+1}=t_{2n}=2$. – Did Dec 15 '13 at 17:24
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    User @imj changed the question by replacing all $\liminf$s with $\lim$s. I've rolledback to the correct version. – Andrés E. Caicedo Dec 15 '13 at 17:52
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    @imj Please do not deface questions, especially to make them fit your answer. – Did Dec 15 '13 at 17:53
  • A different question, with (I think) a different answer: does $\lim \inf (s_m \cdot t_n) = \lim\inf(s_m)\cdot\lim\inf(t_n)$? – TonyK Dec 15 '13 at 17:56
  • my bad, I did not understand the question. I read inf as in infinity, not as in min, and though the OP didn't know how to TeX it. – imj Dec 15 '13 at 17:59
  • @Did : Wouldn't a better (at least less confusing) notation be $\inf\limits_{n\in\mathbb{N}} t_n$ then ? (or just $\inf t_n$ if it's not over natural integers) – imj Dec 15 '13 at 18:10
  • @imj No, these are different notions. – Did Dec 15 '13 at 18:14
  • @Did : after some looking up, $\liminf\limits_{n\to+\infty}$ does indeed mean something else. Again, my apologies, I learned something today. – imj Dec 15 '13 at 18:36
  • The tag ([tag:limit-theorems]) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) – Martin Sleziak Jan 29 '14 at 08:34
  • See also: http://math.stackexchange.com/questions/113121/lim-sup-inequality-limsup-a-n-b-n-leq-limsup-a-n-limsup-b-n and http://math.stackexchange.com/questions/113121/lim-sup-inequality-limsup-a-n-b-n-leq-limsup-a-n-limsup-b-n – Martin Sleziak Jan 29 '14 at 08:35
  • BTW the equality holds if one of the sequences converges: http://math.stackexchange.com/questions/275124/checking-of-a-solution-to-how-to-show-that-lim-sup-a-nb-n-ab – Martin Sleziak Jan 29 '14 at 08:38

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Take a sequence which has liminf -1, and create a new sequence by putting the old sequence as the even terms, and zeros for the odd terms.

Take another sequence with liminf -2, and make a new sequence by putting the old sequence as odd terms and zeros on the even terms.

Both new sequences retain their old liminf, but the term wise product sequence is uniformly zero.

rschwieb
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