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Lim sup problem

  1. Oct 11, 2007 #1
    So I've been trying to figure this problem out for a while now and really am having trouble solving it:

    Show that limsup(sn + tn) ≤ limsup(sn) + limsup(tn), for bounded sequences (sn) and (tn)

    Thus far I've been able to show that limsup(sn) and limsup(tn) are in fact real numbers. Everything in the back of my head screams triangle inequality just looking at the problem, and I'm assuming I need to use the epsilon definition of limits to get that to work, but I'm really stuck getting that far.

    Thanks for any help you guys have to offer.
  2. jcsd
  3. Oct 11, 2007 #2
    wats ur definition of limsup? the monotonic sequence of sup's or supE wer E is the set of subsequential limits of a sequence?

    i found that this actually isn't as simple as it looks.

    Take the latter definition. There is a theorem that says limsup is inside E (sets of limit points are closed is how you get this). Hint: for the LHS, there is a subsequence of sn+tn that approaches it.
  4. Oct 11, 2007 #3
    So forget what I said above - it's leads you to a difficult and rather convoluted proof.

    I found an easier way. We show that any thing greater than the RHS is greater than the LHS, which gives us our theorem.

    Let t>limsuptn, s>limsupsn, then there are only finitely many sn, tn s.t. tn>t, sn>s, so there are only finitely many sn+tn> s+t, so limsup(sn+tn) is bounded above by s+t.

    courtesy of bartle's elementary analysis - slick!
  5. Oct 12, 2007 #4
    Hey, thanks for the input. I actually think I got the solution on my own however, but I'm not sure if there are some shortcuts I can take in my proof (it ended up taking 2 pages...)

    Here's how it goes...if anyone sees glaring errors please point them out to me, I'd prefer that this be as close to perfect as possible, since I need to distribute it to my class.

    Show that limsup(sn + tn) ≤ limsup(sn) + limsup(tn), for bounded sequences (sn) and (tn)

    Sn is said to be bounded and therefore there exists an M in the reals such that |sn| ≤ M for all n. Consider the set {sn : n>N}. For any sn in this set it is obvious that |sn|≤ M also, since it was originally stated that |sn|≤M for all n. Therefore {sn: n>N} is bounded. By the completeness axiom we know that sup{sn: n>N} exists and is real.

    It can be shown similarly that {tn:n>N} is bounded and therefore sup{tn:n>N} exists and is real.

    Now consider the set {sn+tn:n>N}. Since sn is bounded, there exists an M1 such that |sn|≤M1 for all n. Also since tn is bounded there exists an M2 such that |tn|≤M2 for all N. Then |sn|+|tn| ≤ M1+M2 for all n. By the triangle inequality |sn+tn| ≤ |sn|+|tn| therefore by transitivity, we have |sn+tn| ≤ M1+M2 for all n. So {sn+tn: n>N} is bounded. By the completeness axiom, sup {sn+tn: n>N} exists and is real.

    Nw that we know sup{sn:n>N}, sup{tn:n>N}, and sup{sn+tn: n>N} are all real, we must show:
    sup{sn+tn:n>N} ≤ sup{sn:n>N}+sup{tn:n>N}

    We will show this by contradiction.
    Assume then that:

    sup{sn+tn:n>N} > sup{sn:n>N}+sup{tn:n>N}

    This implies that there exists an element a in the set {sn+tn :n>N} such that


    but since a is an element of {sn+tn:n>N}, a = sk+tk where k>N. So we have:

    a=sk+tk > sup{sn:n>N}+sup{tn:n>N}

    0 > sup{sn:n>N}-sk + sup{tn:n>N}-tk

    But sk is an element of {sn:n>N} so sk≤sup{sn:n>N}
    and tk is an element of {tn:n>N} so tk≤sup{tn:n>N}

    So 0≤ sup{sn:n>N}-sk
    and 0≤ sup{tn:n>N}-tk

    So our previous statement is now:
    0>sup{sn:n>N}-sk + sup{tn:n>N}-tk ≥ 0 + 0

    or that:

    which is clearly a contradiction since 0=0. So our initial assumption that sup{sn+tn:n>N} > sup{sn:n>N}+sup{tn:n>N} is incorrect, and therefore we must have

    sup{sn+tn:n>N} ≤ sup{sn:n>N}+sup{tn:n>N}.

    Since {sn+tn:n>N} is bounded it's set A of subsequential limits must be bounded below by inf(sn+tn) and bounded above by sup(sn+tn). We have shown already that (sn+tn) is bounded, so inf and sup are both real numbers by the completeness axiom.Therefore A is bounded and supA is a real number by the completeness axiom. And by theorem 11.7(ii), sup A = limsup{sn+tn:n>N}. So limsup{sn+tn:n>N} is real.

    It is shown similarly that limsup{sn:n:N} and limsup{tn:n>N} are both real. Then by theorem 9.3, limsup{sn:n>N}+limsup{sn:n>N} = lim(sup{sn:n>N}+sup{tn:n>N})

    We have now shown that limsup{sn+tn:n>N} limsup{sn:n>N} and limsup{tn:n>N} all exist and are real.

    We also have shown that sup{sn+tn:n>N} ≤ sup{sn:n>N}+sup{tn:n>N}.

    So by exercise 9.9, which states:
    given N such that sn≤ tn for all n>n, if lims n and lim tn exist, then lim sn ≤ lim tn,

    We can say limsup{sn+tn:n>N} ≤ lim(sup{sn:n>N}+sup{tn:n>N}).
    By theorem 9.3, this is the same as:

    limsup{sn+tn:n>N} ≤ limsup{sn:n>N}+limsup{tn:n>N}.

    Finally, from definition 10.6, we have
    limsup(sn+tn)≤ limsup(sn) + limsup(tn)

    Phew...that was a long one. Thanks to anyone who reads through this for me.
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