Is it true that $limsup a_n geq limsup b_n$ implies $limsup a_n^s_n geq limsup b_n^s_n$ for $a_n, b_n geq 0$ and $s_n > 0$ and increasing?

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Potential first step: If $limsup a_n geq limsup b_n$, along some subsequence and for large enough $n_k$ we have $a_n_k geq b_n_k$. Then, $a_n_k^s_n_k geq b_n_k^s_n_k$, so somehow $limsup a_n_k^s_n_k geq limsup b_n_k^s_n_k$?




sequences-and-series limsup-and-liminf




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  • You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
    – Danny Pak-Keung Chan
    Jul 27 at 1:16










  • We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
    – DanielWainfleet
    Jul 27 at 1:31











  • C_n should have been s_n and that's the increasing sequence
    – user509037
    Jul 27 at 1:50














up vote
0
down vote

favorite












Potential first step: If $limsup a_n geq limsup b_n$, along some subsequence and for large enough $n_k$ we have $a_n_k geq b_n_k$. Then, $a_n_k^s_n_k geq b_n_k^s_n_k$, so somehow $limsup a_n_k^s_n_k geq limsup b_n_k^s_n_k$?




sequences-and-series limsup-and-liminf




share|cite|improve this question





















  • You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
    – Danny Pak-Keung Chan
    Jul 27 at 1:16










  • We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
    – DanielWainfleet
    Jul 27 at 1:31











  • C_n should have been s_n and that's the increasing sequence
    – user509037
    Jul 27 at 1:50












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Potential first step: If $limsup a_n geq limsup b_n$, along some subsequence and for large enough $n_k$ we have $a_n_k geq b_n_k$. Then, $a_n_k^s_n_k geq b_n_k^s_n_k$, so somehow $limsup a_n_k^s_n_k geq limsup b_n_k^s_n_k$?




sequences-and-series limsup-and-liminf




share|cite|improve this question













Potential first step: If $limsup a_n geq limsup b_n$, along some subsequence and for large enough $n_k$ we have $a_n_k geq b_n_k$. Then, $a_n_k^s_n_k geq b_n_k^s_n_k$, so somehow $limsup a_n_k^s_n_k geq limsup b_n_k^s_n_k$?





sequences-and-series limsup-and-liminf





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 1:49
























asked Jul 27 at 0:49









user509037

664




664











  • You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
    – Danny Pak-Keung Chan
    Jul 27 at 1:16










  • We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
    – DanielWainfleet
    Jul 27 at 1:31











  • C_n should have been s_n and that's the increasing sequence
    – user509037
    Jul 27 at 1:50
















  • You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
    – Danny Pak-Keung Chan
    Jul 27 at 1:16










  • We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
    – DanielWainfleet
    Jul 27 at 1:31











  • C_n should have been s_n and that's the increasing sequence
    – user509037
    Jul 27 at 1:50















You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
– Danny Pak-Keung Chan
Jul 27 at 1:16




You may need to rephrase your question. For example, where comes the $s_n$. The word "increasing" refers to which sequence?
– Danny Pak-Keung Chan
Jul 27 at 1:16












We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
– DanielWainfleet
Jul 27 at 1:31





We may have $a_n<b_n$ for every $n$ and still have $limsup a_n=lim sup b_n.$ For example $a_n=1-2^-n$ and $b_n=(1+a_n)/2.$
– DanielWainfleet
Jul 27 at 1:31













C_n should have been s_n and that's the increasing sequence
– user509037
Jul 27 at 1:50




C_n should have been s_n and that's the increasing sequence
– user509037
Jul 27 at 1:50










1 Answer
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Let $a_n = 1-1 over n, b_n = 1+ 1 over n$. Let $s_n = n$.



Then $1 = limsup_n a_n ge limsup_n b_n = 1$, but
$1 over e = limsup_n a_n^s_n < limsup_n b_n^s_n = e$.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Let $a_n = 1-1 over n, b_n = 1+ 1 over n$. Let $s_n = n$.



    Then $1 = limsup_n a_n ge limsup_n b_n = 1$, but
    $1 over e = limsup_n a_n^s_n < limsup_n b_n^s_n = e$.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Let $a_n = 1-1 over n, b_n = 1+ 1 over n$. Let $s_n = n$.



      Then $1 = limsup_n a_n ge limsup_n b_n = 1$, but
      $1 over e = limsup_n a_n^s_n < limsup_n b_n^s_n = e$.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $a_n = 1-1 over n, b_n = 1+ 1 over n$. Let $s_n = n$.



        Then $1 = limsup_n a_n ge limsup_n b_n = 1$, but
        $1 over e = limsup_n a_n^s_n < limsup_n b_n^s_n = e$.






        share|cite|improve this answer













        Let $a_n = 1-1 over n, b_n = 1+ 1 over n$. Let $s_n = n$.



        Then $1 = limsup_n a_n ge limsup_n b_n = 1$, but
        $1 over e = limsup_n a_n^s_n < limsup_n b_n^s_n = e$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 27 at 4:24









        copper.hat

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