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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 n –  Mathematics Sep 9 '12 at 13:26
    
@Mathematics: How shall I prove that? And how will it help in this problem? Please explain. –  Sayantan Sep 9 '12 at 13:33
    
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 contradiction –  Henry Sep 9 '12 at 13:37
    
${\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}} \le \frac{n(\sup s_n)}{n}$ and take $\lim$ on both side. –  Mathematics Sep 9 '12 at 15:22
    
This is a consequence of Stolz-Cesaro theorem, see e.g. here. –  Martin Sleziak Sep 9 '12 at 20:40

1 Answer 1

up vote 11 down vote accepted

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. –  Kasper Feb 7 '13 at 17:53
2  
Where do you take the supremum? –  Davide Giraudo Feb 7 '13 at 19:46
    
@Kasper This only proves the strictly weaker inequality $$\limsup_n\sigma_n\leqslant\sup_ns_n.$$ –  Did Sep 27 at 6:30

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