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

$$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$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 $$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?

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 $\alpha>\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$.

Then there exist infinitely many $N$ such that for all $k\ge0$, $$\alpha>\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$ Since $b_{n+1}<b_n$, we have for $k\ge0$ that $\alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$. Thus for any $m\ge0$, \begin{align} \alpha\sum_{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 $m\to\infty$, \begin{align} -\alpha b_{N-1}&<-a_{N-1} \\ \alpha&>\frac{a_{N-1}}{b_{N-1}}.\end{align}Taking finally $N\to\infty$, we must have $\alpha\ge\limsup_{n\to\infty}\frac{a_n}{b_n}$. Thus we can conclude $$\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\ge\limsup_{n\to\infty}\frac{a_n}{b_n}.$$


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 $\alpha>\limsup_{n\rightarrow}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", 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 $k\ge0$, $\alpha>\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^\prime\ge N$, 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 $N^\prime\ge N$ and $k\ge0$ (this would also make taking the limit as $N^\prime\rightarrow\infty$ 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 $\limsup_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\ge\limsup_{n\rightarrow\infty}\frac{a_n}{b_n}$, but the part $$\limsup_{n\rightarrow\infty}\frac{a_n}{b_n}\ge\liminf_{n\rightarrow\infty}\frac{a_n}{b_n}\ge\liminf_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ 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).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.