$a_n=(-1)^{n-1}, \; s_n=\sum_{i=1}^{n}a_i$ 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 $1$?

  • 2
    $\begingroup$ Consider the $\limsup$ and $\liminf$ of $\frac{s_1+\cdots+s_n}n$. $\endgroup$
    – Math1000
    Feb 5, 2019 at 20:10

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .