Mixing
Stolz-Cesàro $0/0$ case: is $limsup frac{a_n}{b_n}le limsupfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$?
$begingroup$
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
asked Jan 16 at 14:00
LearnerLearner
17510
$endgroup$
add a comment |
$begingroup$
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
asked Jan 16 at 14:00
LearnerLearner
17510
$endgroup$
$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19
add a comment |
$begingroup$
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
asked Jan 16 at 14:00
LearnerLearner
17510
$endgroup$
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 $$limsup_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}gelimsup_{ntoinfty}frac{a_n}{b_n}.$$
real-analysis calculus sequences-and-series proof-verification limsup-and-liminf
real-analysis calculus sequences-and-series proof-verification limsup-and-liminf
asked Jan 16 at 14:00
LearnerLearner
17510
asked Jan 16 at 14:00
LearnerLearner
17510
edited Jan 16 at 15:13
Learner
asked Jan 16 at 14:00
LearnerLearner
17510
asked Jan 16 at 14:00
LearnerLearner
17510
asked Jan 16 at 14:00
LearnerLearner
17510
17510
$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19
add a comment |
$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19
$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19
$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19
add a comment |
1 Answer
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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).
answered Jan 16 at 19:52
ThorgottThorgott
600314
$endgroup$
add a comment |
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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).
answered Jan 16 at 19:52
ThorgottThorgott
600314
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 16 at 19:52
ThorgottThorgott
600314
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 16 at 19:52
ThorgottThorgott
600314
$endgroup$
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).
answered Jan 16 at 19:52
ThorgottThorgott
600314
answered Jan 16 at 19:52
ThorgottThorgott
600314
answered Jan 16 at 19:52
ThorgottThorgott
600314
answered Jan 16 at 19:52
ThorgottThorgott
600314
600314
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/599204/…
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 15:19