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I am thinking about a sequence such that the following holds (noticed > not $\geq$)

$\liminf_{n \to \infty} (a_n + b_n) > \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n$

I am not sure if this is allowed, but I tried doing something like

$a_n = \left \{2....-1,1,-1,1,-1,1 \right \}$

$b_n = \left \{2....1,-1,1,-1,1, -1\right \}$

I know the example doesn't work, but can you actually write down the limsup (the end term) like this? Or is this as erroneous as writing $1/0$?

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Sprite sequences are lemon-liming. –  Asaf Karagila Oct 10 '12 at 7:02
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up vote 2 down vote accepted

No, what you’ve written for $a_n$ and $b_n$ simply doesn’t make sense. The sequences that you’re to construct are ordinary infinite sequences $\langle a_1,a_2,\dots\rangle$ and $\langle b_1,b_2,\dots\rangle$, with no last elements.

HINT: Take $a_n=(-1)^n$, so that $\liminf_na_n=-1$. Can you find another sequence $\langle b_1,b_2,\dots\rangle$ such that $\liminf_nb_n$ is also $-1$, but $\langle a_1+b_1,a_2+b_2,\dots\rangle$ is constant with a value greater than $-1$?

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What do you mean my sequence doesn't have 'last elements'? The only other "b_n" I could think of is something that gives (-1,1,-1,-1,-1) and get constant 0s when I add them up –  sidht Oct 10 '12 at 7:03
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@jak: I’ll rephrase that, since it wasn’t clear. I meant that infinite sequences, which is clearly what we’re talking about here, do not have last elements. Your sequences referred not to the things that you wrote down, but to the $\langle a_1,a_2,\dots\rangle$ and $\langle b_1,b_2,\dots\rangle$ that the problem talked about. –  Brian M. Scott Oct 10 '12 at 7:07
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