Prove limit superior of a sequence is greater than or equal to limit superior of the Cesàro mean of that sequence

Asked Nov 23 '15 at 04:56

Active Nov 23 '15 at 07:24

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The problem is that:

Let $(x_n)$ be a bounded sequence, and $x^* =\limsup(x_n)$. Prove $$\lim\sum_{i=0}^n \frac{x_n}{n}\le x^*.$$

I want to prove it by contradiction.

I start with assume that $\lim\sum_{i=0}^n (x_n)/n> x^*$, then I know $\exists N \in \mathbb N$ (natural number) such that $x_n < \lim \sum_{i=0}^n (x_n)/n$ for all $n \ge N$. "

Then I do not know how to find a contradiction, could anyone help me?

edited Nov 23 '15 at 07:24

asked Nov 23 '15 at 04:56

Belive

3

Are you missing a $\frac{1}{n}$? – Nov 23 '15 at 04:57
Your proof cannot work, because if $x_n$ is decreasing you always have $(\sum_{i=0}^n x_n)/n > x^*$... – Emanuele Paolini Nov 23 '15 at 05:36
@JohnMa Yes, thank you very much – Belive Nov 23 '15 at 05:58
1

@EmanuelePaolini or should i start from the definition of limit superior? – Belive Nov 23 '15 at 05:59
1

It is not even clear to me that $\lim\sum_{i=0}^n\frac{x_n}n$ exists. Do you mean $\limsup\sum_{i=0}^n\frac{x_n}n$? – Hagen von Eitzen Nov 23 '15 at 07:39

Prove limit superior of a sequence is greater than or equal to limit superior of the Cesàro mean of that sequence

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