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## Homework Statement

Consider two bounded sequences {a

_{n}} and {b

_{n}} and put r = liminf(a

_{n})[itex]_{n→\infty}[/itex], s = limsup(b

_{n})[itex]_{n→\infty}[/itex] and t = liminf(a

_{n}+ b

_{n})[itex]_{n→\infty}[/itex]

Show that t ≤ r + s, i.e., that

liminf(a

_{n}+ b

_{n})[itex]_{n→\infty}[/itex] ≤ liminf(a

_{n})[itex]_{n→\infty}[/itex] + limsup(b

_{n})[itex]_{n→\infty}[/itex] .

## 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 a

_{n}≤ r + ε and b

_{n}≤ s + ε

## The Attempt at a Solution

I tried to do a proof by contradiction, so:

Assume that t > r + s

Therefore [itex]\exists[/itex]x s.t. t > x > r + s

[itex]\forall[/itex]ε>0 a

_{n}+ b

_{n}≤ t - ε for finitely many terms

Therefore, let ε = t - x

Therefore a

_{n}+ b

_{n}≤ x for finitely many terms.

However, from previous problem we know that

[itex]\forall[/itex]ε>0

a

_{n}≤ r + ε/2 for infinitely many terms

b

_{n}≤ s + ε/2 for infinitely many terms

Therefore a

_{n}+ b

_{n}≤ r + s + ε

Therefore, let ε = x - r - s

Therefore a

_{n}+ b

_{n}≤ x for infinitely many terms

Therefore there is a contradiction.

So t ≤ r + s.

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