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This is a question from the book Methods of Real Analysis by R. R. Goldberg.

If $(s_n)$ is a sequence of real numbers and if $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.

I don't have any idea how to start working on this problem. Please help. Thanks.

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    @Jack Thanks for the rollback and also for your attention to tags. I agree that (means) is questionable here. Feel free to ping me here or [in chat](https://chat.stackexchange.com/transcript/3740/2017/7/6) if you think that these tag-related comments are no longer needed and should be removed.2017-07-06

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Fix an integer $k$. Let $n\geqslant k$. Then $$\sigma_n=\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j\leqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\sup_{l\geqslant k}s_l.$$ Now take on both sides the limsup when $\color{red}{n\to +\infty}$: we get the wanted result.

Taking $s_n:=(-1)^n$, we can see that the inequality may not be an equality.

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    @MartinSleziak Thanks for pinging.2014-12-03
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Here's a simple solution:

Let $x^{*}_k = \sup_{k \geq n} \{x_n\}$. Then, $x^{*}_k \to \limsup_{n \to \infty} \{x_n\}.$

By a simple fact, a convergent sequence's averages converge to the same limit, so $\sigma_n^{*} = \frac{1}{n} \sum_{j=1}^{n} x^{*}_j \to \limsup_{n \to \infty} \{x_n\}$ as well. Now since $\sigma_n \leq \sigma^{*}_n$, the result follows.