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

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    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] .

    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 an ≤ r + ε and bn ≤ 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 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

    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] ?
  2. jcsd
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