$a_n=O(\frac{1}{n}),\ \frac{\sum_{k=1}^nS_k}{n}$ converges $\Rightarrow S_n$ converges

Question

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.

Answer 1

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.