Supremum

by iarnab_kundu, Jun 16, 2015, 9:01 AM

Let $\{a_n\}$ be a bounded sequence of real numbers.
Then we have $\liminf_{n\to\infty}a_n\le\liminf_{n\to\infty}\dfrac{a_1+\ldots+a_n}{n}\le\limsup_{n\to\infty}\dfrac{a_1+\ldots+a_n}{n}\le\limsup_{n\to\infty} a_n$ .

Let $\limsup_{n\to\infty} a_n=L$ . Then for any $\epsilon>0$ there exists an $N\in\mathbb{N}$ such that $a_n<L+\epsilon$ . Thus we have for $n\ge N$ , $\dfrac{a_1+\ldots+a_n}{n}=\dfrac{C}{n}+(L+\epsilon)\dfrac{n-N+1}{n}\le \dfrac{C}{n}+(L+\epsilon)$ . Where $C=a_1+\ldots+a_{N-1}$ .
Thus $\limsup_{n\to\infty}\dfrac{a_1+\ldots+a_n}{n}\le L+\epsilon$ for any $\epsilon>0$ . Thus $\limsup_{n\to\infty}\dfrac{a_1+\ldots+a_n}{n}\le L$ and hence $\limsup_{n\to\infty}\dfrac{a_1+\ldots+a_n}{n}\le\limsup_{n\to\infty} a_n$ . The rest can be done similarly.

This post has been edited 4 times. Last edited by iarnab_kundu, Jun 16, 2015, 9:34 AM

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Cracking the Nut Open

Supremum

by iarnab_kundu, Jun 16, 2015, 9:01 AM

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