Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.
Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n $ is a monotone decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n = 0 $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.
I saw this interesting question: Calculate $$\lim \limits_{n \to \infty} \left( \sqrt[n+1]{(n+1)!} - \sqrt[n]{n!}\right) $$ The first thing that I noticed is that this sequence is the converse of Stolz-Cesàro theorem for the famous limit $\lim \limits_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$, which in fact, is the correct limit for $\lim \limits_{n \to \infty} \left( \sqrt[n+1]{(n+1)!} - \sqrt[n]{n!}\right) $.
This got me wondering: Is the converse of the Stolz-Cesàro theorem correct? In other words:
Converse of Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and if $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l$.
Converse of Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n $ is a monotone decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n=0$, and if $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l$.
The answer is no! For example let $a_n = (-1)^n, \ b_n =n. $ Then $\lim \limits_{n \to \infty} \frac{a_n}{b_n} =0$, but $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}$ doesn't even exist.
$\color{red}{\textbf{I want to ask:}}$ What are the sufficient conditions for $b_n , a_n$ such that we have a converse to the Stolz-Cesàro theorem? Is there a converse to the Stolz-Cesàro theorem somewhere, or has nobody tried to come up with one?
Why do I think there might be a converse ? The limit $\lim \limits_{n \to \infty} \left( \sqrt[n+1]{(n+1)!} - \sqrt[n]{n!}\right) $ gave me the idea that some sequences have a converse to Stolz-Cesàro theorem, and I remembered that there was a converse for Cauchy limit theorem (which, in fact, is a weaker version of the Stolz-Cesàro theorem in Real numbers ).
Cauchy limit theorem:- If $\lim \limits_{n \to \infty} a_n = a \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{\sum \limits_{k=1}^ n a_k}{n} = a$.
Converse of Cauchy limit theorem:- If $|n(a_n - a_{n-1})|\leq M \in \mathbb{R} \forall n $, and if $\lim \limits_{n \to \infty} \frac{\sum \limits_{k=1}^ n a_k}{n} = a$, then $\lim \limits_{n \to \infty} a_n = a $.
