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Math Help - Arithmetics of superior limits

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    Arithmetics of superior limits

    I uploaded the questions that I have, thank you in advance.
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    Re: Arithmetics of superior limits

    It is really hard to give you an answer. There two reasons: first it appears from your #2 that you do not even understand how limit superiors are defined; second the notations vary widely from text to text.

    That said, let (s_n)\subset\mathbb{R} be a sequence then:
    \lim \,\sup \left( {{s_n}} \right) = {\lim _{N \to \infty }}\sup\left\{ {{s_n}:n > N} \right\}

    I will give some examples. The first two are bounded sequences.

    Let s_n=1-n^{-1} then \lim \,\sup \left( {{s_n}} \right)=1

    Let t_n=n^{-1} then \lim \,\sup \left( {{t_n}} \right)=0. This example maybe where the idea of the tail comes from.

    Let u_n=\begin{cases}n^{-1} &: n\text{ is odd}\\n &: n\text{ is even}\end{cases} then \lim \,\sup \left( {{u_n}} \right)=\infty

    For #1, what do you know about those sequences? Are they bounded?
    Last edited by Plato; December 13th 2013 at 10:45 AM.
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    Re: Arithmetics of superior limits

    So, if I understood it correctly, it's the same idea with the partial limits, in other words we have a sequence, and we have the set S, the set of all partial limits of the sequence and the greatest is equal to the limit of supun, and that is actually the same thing with the expression I wrote, right ? For example the third example that you gave has two subsequences , one converges to infinity, and the other to 0, so S={0, infinity} and infinity is superior therefore we say limsup(un)=infinity (I don't know how to write the sign)

    For the first question, yes , they are bounded.
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    Re: Arithmetics of superior limits

    Quote Originally Posted by davidciprut View Post
    For the first question, yes , they are bounded.
    If each sequence is bounded, then (a_n+b_n) is bounded and the \lim\sup exists.

    Look back at reply #2. Let \tau_N=\sup\{t_n: n>N\}.
    Is it clear that the sequence (\tau_n) is non-increasing and bounded below by \lim\sup (t_n)~?

    Also let \sigma_N=\sup\{s_n: n>N\}..
    Is it also clear that the sequence (\sigma_n) is non-increasing and bounded below by \lim\sup (s_n)~?

    Now let \alpha=\lim\sup (a_n),~\beta=\lim\sup (b_n)~\&~\gamma=\lim\sup (a_n+b_n).
    You want to prove that  \gamma\le\alpha+\beta.
    By contradiction, what happens if \alpha+\beta< \gamma~?
    Last edited by Plato; December 13th 2013 at 12:12 PM.
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    Re: Arithmetics of superior limits

    Shouldn't that be alpha + beta is greater or equal to gamma?
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    Re: Arithmetics of superior limits

    Quote Originally Posted by davidciprut View Post
    Shouldn't that be alpha + beta is greater or equal to gamma?
    YES, and I made the edit.
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    Re: Arithmetics of superior limits

    [deleted]
    Last edited by romsek; December 13th 2013 at 12:47 PM. Reason: answer found
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