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$a_n=(-1)^{n-1}, \; s_n=\sum_{i=1}^{n}a_i$ then find $ \lim_{n\to > \infty}\frac{s_1+s_2+\dots s_n}{n}$

$$s_k=1,\; \text{if k is odd and } s_k=0 \text{ if k is even} $$

Cauchy's theorem for a sequence $(x_n) $ in $R$, we have $\lim\frac{x_1+x_2+\dots x_n}{n}=\lim x_n$

How do I make use of this theorem here when $(s_n)$ is oscillating between $0$ and $1$?

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    $\begingroup$ Consider the $\limsup$ and $\liminf$ of $\frac{s_1+\cdots+s_n}n$. $\endgroup$ – Math1000 Feb 5 at 20:10
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I believe that the theorem you are referring to says:

If a sequence of real numbers $\{x_n\}_{n=1}^\infty$ converges to $x \in \mathbb R$, then $\lim_{n\to \infty} \frac{x_1+x_2+\cdots+x_n}n=x$.


The sequence $\{s_n\}_{n=1}^\infty$ however does not converge as your computations suggest. But if we define $c_n:=\frac{s_1+s_2+\cdots+s_n}n$, then we have $c_n=\frac12$ whenever $n$ is even and $c_n=\frac12+\frac1{2n}$ whenever $n$ is odd. Therefore $\lim_{n \to \infty} c_n=\lim_{n \to \infty}\frac{s_1+s_2+\cdots+s_n}n=\frac12$ as both of these aforementioned subsequences clearly converge to $\frac12$.


The sequence is an example of why the converse of the theorem in question is not true.

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