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

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Homework Statement



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

Show that t ≤ r + s, i.e., that
liminf(an + bn)[itex]_{n→\infty}[/itex] ≤ liminf(an)[itex]_{n→\infty}[/itex] + limsup(bn)[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 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 [itex]\exists[/itex]x s.t. t > x > r + s


[itex]\forall[/itex]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

[itex]\forall[/itex]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

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] ?
 

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