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The inequalities are:

$$\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)$$

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What did you try? –  Did Oct 1 '12 at 9:44
Proofs like this are usually easiest done indirectly: Assume "$>$" holds instead of "$\le$". Then the difference is some $\epsilon>0$, ... –  Hagen von Eitzen Oct 1 '12 at 9:47
See Properties of liminf and lmisup of sequences (and the linked questions and links given there. –  Martin Sleziak Oct 1 '12 at 9:53

1 Answer 1

up vote 5 down vote accepted

Hint: First show that $\limsup$ is subadditive, that is, \[ \limsup(a_n + b_n) \le \limsup a_n + \limsup b_n \] for real sequences $(a_n)$, $(b_n)$. Form this conclude using $-\limsup(-a_n) = \liminf a_n$ that $\liminf$ is superadditive (inequality $\ge$ in the above). Then you can use all this to prove \[ \liminf (a_n + b_n) - \limsup b_n = \liminf (a_n + b_n) + \liminf(-b_n) \le \liminf a_n \] and the other inequality you need.

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