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I know that there does exist a converse of the Stolz - Cesaro theorem when

$\lim_{n \to \infty}\frac{b_n}{b_{n+1}} = B \in \mathbb{R} - {1}$

then $\lim_{n \to \infty}\frac{a_n}{b_n} = L \implies \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L$

reference.

Now it uses the condition that $\lim_{n\rightarrow \infty}\frac{b_{n}}{b_{n+1}}=L \in \mathbb{R}-\{1\}$

But when we see the additive property/arithmetic mean case(reference)(from which the multiplicative property/geometric mean case is derived(reference)), it uses the fact that $b_n = n$ which as we can see, does not satisfy the condition given above. So my question is, does the converse of the aforementioned cases exist? And if yes, can we prove it.

What I mean by the converse of the Arithmetic Mean case is that

If $\lim_{n \to \infty} \frac{x_1 + x_2 + x_3 .. x_n}{n} = L \implies \lim_{n \to \infty}x_n = L$

What I mean by the converse of the Geometric Mean case is that

If $\lim_{n \to \infty} \sqrt[n]{x_1 \cdot x_2 \cdot x_3 .. x_n} = L \implies \lim_{n \to \infty}x_n = L$

OR

If $\lim_{n \to \infty} \sqrt[n]y_n = L \implies \lim_{n \to \infty}\frac{y_{n+1}}{y_n} = L$

P.S. : Sorry if I this has been asked before, I could not find it anywhere on the internet.

Ham Lemon
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    what are you talking about in the arithmetic mean case ? That is still contained in the $\implies $ direction or the usual form of Stolz-Cesaro and has nothing to do with the inverse. – dezdichado Nov 20 '23 at 14:47
  • I have edited the question, can you understand what I am asking now? – Ham Lemon Nov 20 '23 at 15:02
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    the arithmetic mean "case" follows from the usual Cesaro-Stolz, not from the converse so there is nothing to worry about $b_n/b_{n+1}\to 1.$ – dezdichado Nov 20 '23 at 15:06
  • The arithmetic mean (AM) case is proved using $b_n = n$ and if we want to prove the converse of the AM case we could use the same manipulation used to prove the converse of the Stolz - Cesaro (SC). But the converse of the SC is not valid when $\frac{b_{n+1}}{b_n} → 1$ which is used to prove the original AM case. So we cannot use the same manipulation by which we proved the converse of the SC to prove the converse of AM case (as the manipulation only works when $\frac{b_{n+1}}{b_n} does not → 1$). That is why I want to know if there is a proof of the converse of the AM case. – Ham Lemon Nov 20 '23 at 15:16
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    Then write that in the question. What you have there is not the converse of AM case. The Converse would be: $$\lim\limits_{n\to\infty}\dfrac{x_1+x_2+\ldots + x_n}{n}\to L\implies \lim\limits_{n\to\infty} x_n= L,, ?$$ – dezdichado Nov 20 '23 at 15:38
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    the more general context on this is called Tauberian theorems and you will be interested in reading about partial converses to Tauberian theorems. – dezdichado Nov 20 '23 at 15:42
  • I am sorry for the misunderstanding, I have corrected my question. Is it fine now? And thank you for the Tauberian Theorems reference but I have actually tried reading them before and I am still not that qualified to understand to fully understand their language/working/concept, sorry – Ham Lemon Nov 20 '23 at 15:49
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    no sweat. A common sufficient condition for the converse to hold is $x_n = o(\frac 1n).$ This can also be relaxed to $x_n = O(\frac 1n)$ if we assume positivity of the sequence etc. There are probably more but in general there is no catch-all converse as far as I know. – dezdichado Nov 20 '23 at 16:11
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    Ok, I understand some of what you said, I will do a further study on what you said. Thank you for that – Ham Lemon Nov 20 '23 at 17:18

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