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

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 $$0 for all $$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 $$b_{n+1}, we have for $$k\ge0$$ that $$\alpha(b_{N+k}-b_{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})&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 $$\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\ge\limsup_{n\to\infty}\frac{a_n}{b_n}.$$

• The assumption $$0 is an additional assumption (or rather $$(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, 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).