Stolz-Cesàro $0/0$ case: is $limsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$?












3












$begingroup$


The general form of Stolz-Cesaro $infty/infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy



$$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$Does the same hold for the $0/0$ case? That is, is it true that if $lim a_n=lim b_n=0$ and $b_n$ is strictly monotone, then $$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?



EDIT: Here's my attempt, please any feedback is appreciated.



I tried with the $limsup$, assuming $0<b_{n+1}<b_n$ for all $n$. Suppose $alpha>limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$.



Then there exist infinitely many $N$ such that for all $kge0$, $$alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$
Since $b_{n+1}<b_n$, we have for $kge0$ that $alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$.
Thus for any $mge0$, begin{align} alphasum_{k=0}^m(b_{N+k}-b_{N+k-1}) &< sum_{k=0}^m(a_{N+k}-a_{N+k-1}) \ alpha(b_{N+m}-b_{N-1})&<a_{N+m}-a_{n-1}end{align}and taking $mtoinfty$, begin{align} -alpha b_{N-1}&<-a_{N-1} \ alpha&>frac{a_{N-1}}{b_{N-1}}.end{align}Taking finally $Ntoinfty$, we must have $alphagelimsup_{ntoinfty}frac{a_n}{b_n}$. Thus we can conclude $$limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{ntoinfty}frac{a_n}{b_n}.$$






real-analysis calculus sequences-and-series proof-verification limsup-and-liminf







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  • $begingroup$
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    $endgroup$
    – Mohammad Zuhair Khan
    Jan 16 at 15:19
















3












$begingroup$


The general form of Stolz-Cesaro $infty/infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy



$$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$Does the same hold for the $0/0$ case? That is, is it true that if $lim a_n=lim b_n=0$ and $b_n$ is strictly monotone, then $$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?



EDIT: Here's my attempt, please any feedback is appreciated.



I tried with the $limsup$, assuming $0<b_{n+1}<b_n$ for all $n$. Suppose $alpha>limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$.



Then there exist infinitely many $N$ such that for all $kge0$, $$alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$
Since $b_{n+1}<b_n$, we have for $kge0$ that $alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$.
Thus for any $mge0$, begin{align} alphasum_{k=0}^m(b_{N+k}-b_{N+k-1}) &< sum_{k=0}^m(a_{N+k}-a_{N+k-1}) \ alpha(b_{N+m}-b_{N-1})&<a_{N+m}-a_{n-1}end{align}and taking $mtoinfty$, begin{align} -alpha b_{N-1}&<-a_{N-1} \ alpha&>frac{a_{N-1}}{b_{N-1}}.end{align}Taking finally $Ntoinfty$, we must have $alphagelimsup_{ntoinfty}frac{a_n}{b_n}$. Thus we can conclude $$limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{ntoinfty}frac{a_n}{b_n}.$$






real-analysis calculus sequences-and-series proof-verification limsup-and-liminf







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$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/599204/…
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 16 at 15:19














3












3








3





$begingroup$


The general form of Stolz-Cesaro $infty/infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy



$$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$Does the same hold for the $0/0$ case? That is, is it true that if $lim a_n=lim b_n=0$ and $b_n$ is strictly monotone, then $$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?



EDIT: Here's my attempt, please any feedback is appreciated.



I tried with the $limsup$, assuming $0<b_{n+1}<b_n$ for all $n$. Suppose $alpha>limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$.



Then there exist infinitely many $N$ such that for all $kge0$, $$alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$
Since $b_{n+1}<b_n$, we have for $kge0$ that $alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$.
Thus for any $mge0$, begin{align} alphasum_{k=0}^m(b_{N+k}-b_{N+k-1}) &< sum_{k=0}^m(a_{N+k}-a_{N+k-1}) \ alpha(b_{N+m}-b_{N-1})&<a_{N+m}-a_{n-1}end{align}and taking $mtoinfty$, begin{align} -alpha b_{N-1}&<-a_{N-1} \ alpha&>frac{a_{N-1}}{b_{N-1}}.end{align}Taking finally $Ntoinfty$, we must have $alphagelimsup_{ntoinfty}frac{a_n}{b_n}$. Thus we can conclude $$limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{ntoinfty}frac{a_n}{b_n}.$$






real-analysis calculus sequences-and-series proof-verification limsup-and-liminf







share|cite|improve this question











$endgroup$




The general form of Stolz-Cesaro $infty/infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy



$$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$Does the same hold for the $0/0$ case? That is, is it true that if $lim a_n=lim b_n=0$ and $b_n$ is strictly monotone, then $$liminffrac{a_{n+1}-a_n}{b_{n+1}-b_n}leliminffrac{a_n}{b_n}lelimsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?



EDIT: Here's my attempt, please any feedback is appreciated.



I tried with the $limsup$, assuming $0<b_{n+1}<b_n$ for all $n$. Suppose $alpha>limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$.



Then there exist infinitely many $N$ such that for all $kge0$, $$alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$
Since $b_{n+1}<b_n$, we have for $kge0$ that $alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$.
Thus for any $mge0$, begin{align} alphasum_{k=0}^m(b_{N+k}-b_{N+k-1}) &< sum_{k=0}^m(a_{N+k}-a_{N+k-1}) \ alpha(b_{N+m}-b_{N-1})&<a_{N+m}-a_{n-1}end{align}and taking $mtoinfty$, begin{align} -alpha b_{N-1}&<-a_{N-1} \ alpha&>frac{a_{N-1}}{b_{N-1}}.end{align}Taking finally $Ntoinfty$, we must have $alphagelimsup_{ntoinfty}frac{a_n}{b_n}$. Thus we can conclude $$limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{ntoinfty}frac{a_n}{b_n}.$$






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edited Jan 16 at 15:13







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$begingroup$

This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).




  • The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.


  • It might be clearer to write "Let $alpha>...$ be arbitrary" to emphasize that you are talking about any such $alpha$.


  • In particular, when you say "Suppose $alpha>limsup_{nrightarrow}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you assume it is real, because the inequality is trivial when the $limsup$ is infinite.

  • The formulation "There exist infinitely many $N$ such that for all $kge0$, $alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^primege N$, so it would be clearer to say something like $alpha>frac{a_{N^prime+k}-a_{N^prime+k-1}}{b_{N^prime+k}-b_{N^prime+k-1}}$ for all $N^primege N$ and $kge0$ (this would also make taking the limit as $N^primerightarrowinfty$ clearer).

  • In the next line it becomes apparent that choosing the symbol $alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

  • Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $lim a_n=lim b_n$), so the inequalities $-alpha b_{N-1}<-a_{N-1}$ and $alpha>frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

  • Lastly, you have proven $limsup_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{nrightarrowinfty}frac{a_n}{b_n}$, but the part
    $$limsup_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
    is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $liminf$ inequality follows directly from the $limsup$ inequality (and maybe how so).






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    $begingroup$

    This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).




    • The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.


    • It might be clearer to write "Let $alpha>...$ be arbitrary" to emphasize that you are talking about any such $alpha$.


    • In particular, when you say "Suppose $alpha>limsup_{nrightarrow}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you assume it is real, because the inequality is trivial when the $limsup$ is infinite.

    • The formulation "There exist infinitely many $N$ such that for all $kge0$, $alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^primege N$, so it would be clearer to say something like $alpha>frac{a_{N^prime+k}-a_{N^prime+k-1}}{b_{N^prime+k}-b_{N^prime+k-1}}$ for all $N^primege N$ and $kge0$ (this would also make taking the limit as $N^primerightarrowinfty$ clearer).

    • In the next line it becomes apparent that choosing the symbol $alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

    • Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $lim a_n=lim b_n$), so the inequalities $-alpha b_{N-1}<-a_{N-1}$ and $alpha>frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

    • Lastly, you have proven $limsup_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{nrightarrowinfty}frac{a_n}{b_n}$, but the part
      $$limsup_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
      is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $liminf$ inequality follows directly from the $limsup$ inequality (and maybe how so).






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).




      • The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.


      • It might be clearer to write "Let $alpha>...$ be arbitrary" to emphasize that you are talking about any such $alpha$.


      • In particular, when you say "Suppose $alpha>limsup_{nrightarrow}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you assume it is real, because the inequality is trivial when the $limsup$ is infinite.

      • The formulation "There exist infinitely many $N$ such that for all $kge0$, $alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^primege N$, so it would be clearer to say something like $alpha>frac{a_{N^prime+k}-a_{N^prime+k-1}}{b_{N^prime+k}-b_{N^prime+k-1}}$ for all $N^primege N$ and $kge0$ (this would also make taking the limit as $N^primerightarrowinfty$ clearer).

      • In the next line it becomes apparent that choosing the symbol $alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

      • Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $lim a_n=lim b_n$), so the inequalities $-alpha b_{N-1}<-a_{N-1}$ and $alpha>frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

      • Lastly, you have proven $limsup_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{nrightarrowinfty}frac{a_n}{b_n}$, but the part
        $$limsup_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
        is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $liminf$ inequality follows directly from the $limsup$ inequality (and maybe how so).






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).




        • The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.


        • It might be clearer to write "Let $alpha>...$ be arbitrary" to emphasize that you are talking about any such $alpha$.


        • In particular, when you say "Suppose $alpha>limsup_{nrightarrow}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you assume it is real, because the inequality is trivial when the $limsup$ is infinite.

        • The formulation "There exist infinitely many $N$ such that for all $kge0$, $alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^primege N$, so it would be clearer to say something like $alpha>frac{a_{N^prime+k}-a_{N^prime+k-1}}{b_{N^prime+k}-b_{N^prime+k-1}}$ for all $N^primege N$ and $kge0$ (this would also make taking the limit as $N^primerightarrowinfty$ clearer).

        • In the next line it becomes apparent that choosing the symbol $alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

        • Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $lim a_n=lim b_n$), so the inequalities $-alpha b_{N-1}<-a_{N-1}$ and $alpha>frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

        • Lastly, you have proven $limsup_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{nrightarrowinfty}frac{a_n}{b_n}$, but the part
          $$limsup_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
          is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $liminf$ inequality follows directly from the $limsup$ inequality (and maybe how so).






        share|cite|improve this answer









        $endgroup$



        This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).




        • The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.


        • It might be clearer to write "Let $alpha>...$ be arbitrary" to emphasize that you are talking about any such $alpha$.


        • In particular, when you say "Suppose $alpha>limsup_{nrightarrow}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you assume it is real, because the inequality is trivial when the $limsup$ is infinite.

        • The formulation "There exist infinitely many $N$ such that for all $kge0$, $alpha>frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^primege N$, so it would be clearer to say something like $alpha>frac{a_{N^prime+k}-a_{N^prime+k-1}}{b_{N^prime+k}-b_{N^prime+k-1}}$ for all $N^primege N$ and $kge0$ (this would also make taking the limit as $N^primerightarrowinfty$ clearer).

        • In the next line it becomes apparent that choosing the symbol $alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

        • Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $lim a_n=lim b_n$), so the inequalities $-alpha b_{N-1}<-a_{N-1}$ and $alpha>frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

        • Lastly, you have proven $limsup_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{nrightarrowinfty}frac{a_n}{b_n}$, but the part
          $$limsup_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_n}{b_n}geliminf_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
          is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $liminf$ inequality follows directly from the $limsup$ inequality (and maybe how so).







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        answered Jan 16 at 19:52









        ThorgottThorgott

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