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Question: Let $a_n$ be a sequence of real numbers. Is it true that for every $\varepsilon > 0$, if

$$\left \lvert \frac{1}{N} \left ( \sum_{n=0}^{N-1} a_n \right )\right \rvert < \frac{1}{N^{1+\varepsilon}}$$

for all $N \in \mathbb Z_+$, then $a_n$ converges to $0$?

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    $\begingroup$ This implies $\sum a_n$ converges (to $0$) and so $a_N = \sum_{i=0}^{N}a_n -\sum_{i=0}^{N-1}a_n$ must tend to $0$ (at rate at least $N^{-\varepsilon}$). $\endgroup$ Commented Apr 23, 2023 at 19:38
  • $\begingroup$ Ah right, this is indeed trivial as stated.. $\endgroup$
    – Nate River
    Commented Apr 23, 2023 at 19:43

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Multiply by $N$ to get that

$$\left \lvert \sum_{n=0}^{N-1} a_n \right \rvert < \frac{1}{N^{\varepsilon}}$$

Therefore $\sum_{n=0}^\infty a_n=0$ and $\lim_{n\to\infty} a_n=0$.

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