Suppose $\{a_n\}_{n=1}^\infty$ is non-increasing sequence that converges to $0$. For convenience let $a_1 \leq 1$. Suppose $S_n = \sum_{k=1}^n a_k$. Is it possible to show that $\exists \alpha \gneq 0$ s.t. $S_n \in O(n^{1-\alpha})$? If not, can it be shown that $S_n \in o(n)$?
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It's always true if you allow $\alpha = 0$, but not if you restrict to $\alpha > 0$. – Antonio Vargas Dec 11 '16 at 14:59
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You have $S_n \in o(n)$, but it can be something like $\Theta(n/\log n)$ or so. – Daniel Fischer Dec 11 '16 at 15:02
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@DanielFischer you got me. Can you please tell me how to prove $S_n \in o(n)$? – aroyc Dec 11 '16 at 15:05
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2You don't even need monotonicity. If a sequence converges, the sequence of Cesàro means converges to the same limit. Here, $S_n \in o(n)$ precisely means that the Cesàro means converge to $0$. – Daniel Fischer Dec 11 '16 at 15:08
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See e.g. here. – Daniel Fischer Dec 11 '16 at 15:11
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@DanielFischer Thanks man! – aroyc Dec 11 '16 at 15:30