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I need help to prove the question or find a counterexample. I think that was true. Proof idea: Since $\lim_{k\to \infty} a_k=0$, given $\epsilon >0$ there is $k_0\in\mathbb{N}$ such that $a_k < \epsilon$ for all $k > k_0$. Then for $n>k_0$, $$ \frac{1}{n}\sum_{k=0}^n a_k =\frac{1}{n}\sum_{k=0}^{k_0} a_k+\frac{1}{n}\sum_{k=k_0+1}^n a_k <\frac{1}{n}\sum_{k=0}^{k_0-1} a_k+\frac{n-k_0}{n} \epsilon $$ so $$ \lim_{n\to \infty} \frac{1}{n}\sum_{k=0}^n a_k <\epsilon, $$ and hence the proof is completed.

Emanuel Ramadori
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  • Assume the average value is nonzero. – Wakem Dec 13 '21 at 18:23
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    You want to use $\leq$ instead of $<$. Note that if a function is nonnegative its limit is not necessarily nonnegative. If you really want to be rigorous you can show the $\limsup$ is less than or equal to $\epsilon$, but your work is basically doing the right thing. – Michael Dec 13 '21 at 18:23
  • The idea is basically correct, but you're mixing up $\varepsilon$-$\delta_{\varepsilon}$ definition and the use of the comparison theorem; I would use the fact that, since for any fixed $\varepsilon>0$ and any fixed $k_0 \in \mathbb{N}$ the sequence $(a_j)_j =\frac{j-k_0}{j}\varepsilon$ tends to $\varepsilon$ as $j \to \infty$, you can get the estimation you've already obtained. – Bernkastel Dec 13 '21 at 18:24
  • i forget to assume that $a_k\geq0$. thanks for the answer – Emanuel Ramadori Dec 13 '21 at 18:27
  • @EmanuelRamadori: You're welcome, it is preferred to edit your questions instead of making corrections in the comments; you can use the option "edit" under your question. – Bernkastel Dec 13 '21 at 18:29
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    Thanks for the advice, I already updated the question – Emanuel Ramadori Dec 13 '21 at 18:32

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If $(x_n) \to 0$ then $\lim_{n \to \infty}\frac{x_1+x_2+x_3+...+x_n}{n}=0$

Please see this to get a proof of Cesaro convergence

Sourav Ghosh
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