2
$\begingroup$

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 $a_n = O(c_n)$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.