Let $(a_n)$ be a sequence of real numbers such that $n|a_n|\le M\ \forall n$ for some positive number $M$. We define the sequences $S_n=\sum_{k=1}^n a_k$ and $\sigma_n=\frac{\sum_{k=1}^nS_k}{n}\ \forall n$. Then we have the following result:
$(\sigma_n)$ converges $\Rightarrow (S_n)$ converges.
I think I have seen this result before and I think it has a specific name, but I can't remember it. Does anyone know its name ? And could someone provide me with a proof or some tips on how to prove it ?
Thank you in advance for your help.
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Tengen
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The converse of this statement follows directly from Stolz-Cesàro theorem – Peter Foreman Sep 21 '19 at 17:27
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@PetwrForeman Isn't the implication in the other direction? Edit: I missed the "converse" in your comment. – Botond Sep 21 '19 at 17:28
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@PeterForeman Yes I agree that the converse is a consequence of Stolz-Cesaro theorem. But I'm looking for a proof for the direct implication. – Tengen Sep 21 '19 at 17:39
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The other answers are right, but the OP’s implication is actually correct.
Cesaro summable implies Abel summable shows that in your situation, the sequence $\sum{S_n-S_{n-1}}$ is Cesaro-summable, so is Abel summable.
Now, $S_n-S_{n-1}=O(1/n)$, so by the examples of Tauber theorems here ( https://en.m.wikipedia.org/wiki/Abelian_and_Tauberian_theorems ) $\sum{S_n-S_{n-1}}$ converges.
Aphelli
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