For a sequence $a_n = O(n^{-1/2})$ as $n\to\infty$, consider the corresponding Cesàro means $b_n = \frac{1}{n} \sum_{j=1}^n a_j$.

Is it possible to derive the rate of convergence for the sequence $b_n$?

What about the general case $a_n = O(c_n)$?


After some further searching, I believe I have the answer at least for $a_n = \mathcal O(n^r)$ and $r\in\mathbb R$.

By assumptions, there exists $c>0$ such that for all $n$ large enough $\left\vert a_n \right\vert \leq c n^{r}$. Now, $$ \left\vert b_n \right\vert \leq \frac{1}{n}\sum_{j=1}^n \left\vert a_j \right\vert \leq \frac{c}{n} \sum_{j=1}^n j^{r}, $$ and it is enough to bound the last sum. This can be done by Euler's summation formula (cf Apostol, 1976, Theorem 3.2, or here), which provides $$ b_n = \mathcal O(n^r) \quad \mbox{ for }r> -1,$$ $$ b_n = \mathcal O(n^{-1}) \quad \mbox{ for }r<-1,$$ and $$ b_n = \mathcal O(\log(n)/n) \quad \mbox{ for }r=-1.$$

The general case seems to be more involved.


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