(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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] .

2. Relevant 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

3. 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] ?

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

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