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Lim sup(an+bn) less than or equal to

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    If limsup(an) and limsup(bn) are finite, prove that limsup(an+bn) [tex]\leq[/tex] limsup(an) + limsup(bn).


    2. Relevant equations



    3. The attempt at a solution
    My proof seems a bit short, so if someone could please reassure me this is a valid proof, thanks in advance.

    Proof: Assuming an and bn are bounded sequence. Let a > limsup(an) and b > limsup(bn). Then a+b > an+bn for all but finitely many n's. This implies that a+b [tex]\geq[/tex] limsup(an+bn). Since this hold for any a [tex]\geq[/tex] limsup(an) and any b > limsup(bn), this implies limsup(an+bn) [tex]\leq[/tex] limsup(an) + limsup(bn). QED
     
  2. jcsd
  3. Feb 18, 2010 #2

    lanedance

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    looks reasonable to me (though i am rusty at this)

    so just to sum up

    an > limsup(an), bn > limsup(bn) for only finitley many n

    an + bn > limsup(an + bn) for only finitley many n

    which implies
    limsup(an + bn) <= limsup(an) + limsup(bn)
     
    Last edited: Feb 18, 2010
  4. Feb 18, 2010 #3
    thanks for the input, unfortunately my professor refused to tell me if i was right.
     
  5. Feb 20, 2010 #4


    Your first sentence is too strong an assumption. E.g., let x_n = -n, if n even, 1 otherwise. Then lim sup x_n = 1, but the sequence {x_n} is not bounded.

    Edit: oops, I just looked at your proof, you didn't even use the boundedness...
     
    Last edited: Feb 20, 2010
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