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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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    well, you may consider that $\sup \sigma_n \leq \sup \ s_n$ for all n2012-09-09
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    @Mathematics: How shall I prove that? And how will it help in this problem? Please explain.2012-09-09
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    Assume the opposite, that there is some $k$ with $\operatorname{lim sup} s_n \lt k \lt \operatorname{lim sup}\sigma_n$, and find a contradiction2012-09-09
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    ${\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}} \le \frac{n(\sup s_n)}{n}$ and take $\lim$ on both side.2012-09-09
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    This is a consequence of Stolz-Cesaro theorem, see e.g. [here](http://math.stackexchange.com/a/100542/8297).2012-09-09
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    @Mathematics This only proves the strictly weaker inequality $$\limsup_n\sigma_n\leqslant\sup_ns_n.$$2014-09-27
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    @Jack Why do you think that ([tag:limsup]) and ([tag:inequality]) tag are not suitable for this question?2017-07-06
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    @MartinSleziak: Rolled back. Those two should be relevant while `means` is not.2017-07-06
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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

2 Answers 2

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Fix an integer $k$. Let $n\geqslant k$. Then $$\sigma_n=\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j\leqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\sup_{l\geqslant k}s_l\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l.$$ 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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    Can you prove it like this: ${\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}} \le \frac{n(\sup s_n)}{n}$ and take lim on both sides ? This seems easier to me, but it seems to easy to be right.2013-02-07
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    Where do you take the supremum?2013-02-07
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    @Kasper This only proves the strictly weaker inequality $$\limsup_n\sigma_n\leqslant\sup_ns_n.$$2014-09-27
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    [Another question](http://math.stackexchange.com/questions/1049869/limsup-of-average-smaller-than-limsup) asking for clarification for this answer was posted. (I tried to answer it, but in any case I thought it would be polite to leave you a ping. Maybe you will have something to say to the OP of the other question, too.)2014-12-03
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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.