$\lim\inf a_n+\lim\sup b_n\le\lim\sup(a_n+b_n)\le\lim\sup a_n+\lim\sup b_n$

For the right inequality, I assume $A=\lim\sup a_n, B = \lim\sup B_n$. Hence for any $\varepsilon$, there exists $n_0$ such that $a_{n}<A+\varepsilon$ for all $n\ge n_0$. Similarly for $b_n$. Hence for any $\varepsilon$, there exists $n_0$ such that $a_n+b_n<A+B+\varepsilon$ for all $n\ge n_0$. Therefore $\lim\sup(a_n+b_n)\leq A+B$.

Now for the left inequality, assume $A=\lim\inf a_n, B = \lim\sup B_n$. How can this be compared to $\lim\sup(a_n+b_n)$?

  • $\begingroup$ possible duplicate of How to prove these inequalities in real analysis? $\endgroup$
    – Evan
    Aug 25, 2013 at 3:01
  • $\begingroup$ Just realized possibly something to address. The right inequality is good. A lower bound on limsup, $\limsup a_n \geq c$ means that $a_n \geq c+\epsilon$ infinitely often for some $\epsilon$. So I suppose you need to show that for some $\epsilon$, $a_n + b_n > A + B + \epsilon$ infinitely often (or if you like, for every $N$ there is some $n > N$ for which this occurs). $\endgroup$
    – Evan
    Aug 25, 2013 at 3:15

1 Answer 1


If you chase two links down you have a possible solution, though perhaps you won't want to look at that. Also I thought of a different approach along the lines of my comment that I want to share, but just with hints:

Goal: For some $\epsilon$, $a_n + b_n > A+B+\epsilon$ infinitely often.

  1. Definition of $B$: There is some $\epsilon$ with $b_n > B+\epsilon$ for infinitely many $n$.
  2. Use definition of $A$ to get that for any $\delta$, $a_n > A - \delta$ for sufficiently large $n$. Choose $\delta$ appropriately.
  • $\begingroup$ Thanks, I think you mean $b_n>b-\epsilon$. $\endgroup$
    – PJ Miller
    Aug 25, 2013 at 12:30
  • $\begingroup$ @PJMiller bn is for limsup using the lower bound characterization, so has to be + $\endgroup$
    – Evan
    Aug 25, 2013 at 15:16
  • $\begingroup$ To see the lower bound characterization you need to write the negation of your characterization for limsup an < a $\endgroup$
    – Evan
    Aug 25, 2013 at 15:22

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