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Math Help - i cant find the basic lim inf=lim sup convergence proof

  1. #1
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    i cant find the basic lim inf=lim sup convergence proof

    i look every where for that basic proof
    that if
    lim inf An=lim sup An then the sequence converges

    ??
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by transgalactic View Post
    i look every where for that basic proof
    that if
    lim inf An=lim sup An then the sequence converges

    ??
    In my lecture notes, it is denoted as a definition.

    That is :
    (A_n) converges if \limsup A_n=\liminf A_n
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  3. #3
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    i have to proove this definition
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    i look every where for that basic proof
    that if
    lim inf An=lim sup An then the sequence converges

    ??
    Let S be the set of all subsequential limits of \left\{a_n\right\}. Also let \inf S=\sup S=\xi. So it is clear that if x\in S then \xi\leqslant x \leqslant \xi\implies x=\xi. Now let us define a subsequence of \left\{a_n\right\} as \left\{a_{\varphi}\right\} where \varphi:\mathbb{N}\mapsto\mathbb{N} and \varphi is strictly increasing. So since \left\{a_{\varphi}\right\}\to\xi~\forall\varphi it is clear that there exists a N such that N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon.Now based on the above definition we may say that as soon as \lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon which says exactly that not only does a_n converge, but it converges to \xi.

    That is how I would do it. Is that acceptable?

    You may find this thread helpful http://www.mathhelpforum.com/math-he...-analysis.html
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Let S be the set of all subsequential limits of \left\{a_n\right\}. Also let \inf S=\sup S=\xi. So it is clear that if x\in S then \xi\leqslant x \leqslant \xi\implies x=\xi. Now let us define a subsequence of \left\{a_n\right\} as \left\{a_{\varphi}\right\} where \varphi:\mathbb{N}\mapsto\mathbb{N} and \varphi is strictly increasing. So since \left\{a_{\varphi}\right\}\to\xi~\forall\varphi it is clear that there exists a N such that N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon.Now based on the above definition we may say that as soon as \lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon which says exactly that not only does a_n converge, but it converges to \xi.

    That is how I would do it. Is that acceptable?

    You may find this thread helpful http://www.mathhelpforum.com/math-he...-analysis.html
    i don't like this proof. i don't think it does the job. first, i think you mean to use limsup and liminf where you use sup and inf. secondly, your proof suggests that the sequence is constant, which of course, is not true for convergent sequences in general.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    i think you mean to use limsup and liminf where you use sup and inf.
    No I don't \limsup a_n=\sup\left(S\right) where S is the set of all subsequential limits. The same goes for \liminf a_n=\inf\left(S\right)
    secondly, your proof suggests that the sequence is constant, which of course, is not true for convergent sequences in general.
    I'm not quite sure what you mean by this? Do you mean the f maps to a single value kind of constant or some other kind...either way I don't think I assumed anything
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    No I don't \limsup a_n=\sup\left(S\right) where S is the set of all subsequential limits. The same goes for \liminf a_n=\inf\left(S\right)

    I'm not quite sure what you mean by this? Do you mean the f maps to a single value kind of constant or some other kind...either way I don't think I assumed anything
    oh, ok, i missed your definition of S... sorry

    i will look over the proof again later. i don't have time now
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