# Prove that liming(a_n + b_n) ≤ liminf a_n + limsup b_n

## Homework Statement

Consider two bounded sequences {an} and {bn} and put r = liminf(an)$_{n→\infty}$, s = limsup(bn)$_{n→\infty}$ and t = liminf(an + bn)$_{n→\infty}$

Show that t ≤ r + s, i.e., that
liminf(an + bn)$_{n→\infty}$ ≤ liminf(an)$_{n→\infty}$ + limsup(bn)$_{n→\infty}$ .

## Homework Equations

In the part of the problem before we needed to show that for every ε>0 there are an infinite number of integers n such that an ≤ r + ε and bn ≤ s + ε

## The Attempt at a Solution

I tried to do a proof by contradiction, so:

Assume that t > r + s

Therefore $\exists$x s.t. t > x > r + s

$\forall$ε>0 an + bn ≤ t - ε for finitely many terms

Therefore, let ε = t - x

Therefore an + bn ≤ x for finitely many terms.

However, from previous problem we know that

$\forall$ε>0
an ≤ r + ε/2 for infinitely many terms
bn ≤ s + ε/2 for infinitely many terms

Therefore an + bn ≤ r + s + ε

Therefore, let ε = x - r - s

Therefore an + bn ≤ x for infinitely many terms

So t ≤ r + s.

I think that makes sense, but does that cover all possible values for r and s? What if they are ±$\infty$ ?