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How would you prove/show the cesaro mean $\lim_{n \to \infty}\left(\frac{1}{n}\sum_{k=1}^n a_k\right)$ of an aternating sequence such as: $$a_k = \begin{cases} 1 & k\equiv 0\mod 3 \\ 0 & k\not\equiv 0 \mod 3\end{cases}$$ Intuitively the mean would be $\frac{1}{3}$. But I am having a hard time conceiving a formal proof. How would you approach this?

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Hint. Start by showing that for any $n\geq 1$, $$\frac{n}{3}-1\leq \sum_{k=1}^n a_k \leq \frac{n}{3}.$$

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  • $\begingroup$ shouldn't the middle term be divided by $n$? Or am I totally missunderstanding the hint? $\endgroup$ – Alex Q Oct 20 '16 at 5:18
  • $\begingroup$ @Alex Q No. If you divide the middle term by $n$, you should divide by $n$ also the term on the left and the term on the right. $\endgroup$ – Robert Z Oct 20 '16 at 5:52

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