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I am aware of the Caesaro convergence, if $a_n\to a$, then $\sum_{i=1}^n a_i/n\to a$ as well, as is discussed here and here.

I am wondering about a generalization of this result. Can we say anything about the convergence of $\sum_{i=1}^na_i/n^p$ where $p\in\mathbb{R}$? In particular, what if $p\in(0,1)$?

For example, does $\sum_{i=1}^na_i/\sqrt{n}$ converge as well? If so, can we say to what, in terms of $a$?

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    Hint: $n^p = n \cdot n^{p-1}$, so as $a_n \to a$, then $\sum_{i=1}^n \frac{a_i}{n^p} = n^{1-p} \sum_{i=1}^n \frac{a_i}{n}$ – Presage Sep 21 '19 at 02:41
  • Does this imply that the sequence diverges, because if $p\in(0,1)$, $n^{1-p}$ diverges? –  Sep 21 '19 at 02:44
  • If $a \neq 0$ it clearly diverges. If $a_n \to 0$ we must know something about the rate of convergence – Presage Sep 21 '19 at 02:45
  • Could you elaborate about what you mean by the rate of convergence? –  Sep 21 '19 at 02:46
  • $a_n = \frac{1}{n^2}$ converges to $0$ much "stronger" than $b_n = \frac{1}{\sqrt{n}}$ for example. In the first case, $\sum_{k=1}^n a_k= \sum_{k=1}^n \frac{1}{k^2} \le 1+ \int_1^n \frac{1}{x^2}dx = 2 - \frac{1}{n}$, then for any $p>0$ we have $\frac{1}{n^p} \sum_{k=1}^n a_k \to 0$, but for $b_n$, we have : $\sum_{k=1}^n b_k \ge \int_1^n \frac{dx}{\sqrt{x}} = 2\sqrt{n} - 2$, so $\frac{1}{\sqrt{n}}\sum_{k=1}^n b_k$ cannot converge to $0$. – Presage Sep 21 '19 at 02:55
  • ah, okay. Is there anyway to better characterize this sort of "strong" convergence? –  Sep 21 '19 at 02:57
  • Let us continue this discussion in chat. –  Sep 21 '19 at 18:18

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