i look every where for that basic proof

that if

lim inf An=lim sup An then the sequence converges

??

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- Dec 13th 2008, 02:08 AMtransgalactici cant find the basic lim inf=lim sup convergence proof
i look every where for that basic proof

that if

lim inf An=lim sup An then the sequence converges

?? - Dec 13th 2008, 02:32 AMMoo
- Dec 13th 2008, 02:39 AMtransgalactic
i have to proove this definition

- Dec 13th 2008, 09:39 AMMathstud28
Let $\displaystyle S$ be the set of all subsequential limits of $\displaystyle \left\{a_n\right\}$. Also let $\displaystyle \inf S=\sup S=\xi$. So it is clear that if $\displaystyle x\in S$ then $\displaystyle \xi\leqslant x \leqslant \xi\implies x=\xi$. Now let us define a subsequence of $\displaystyle \left\{a_n\right\}$ as $\displaystyle \left\{a_{\varphi}\right\}$ where $\displaystyle \varphi:\mathbb{N}\mapsto\mathbb{N}$ and $\displaystyle \varphi$ is strictly increasing. So since $\displaystyle \left\{a_{\varphi}\right\}\to\xi~\forall\varphi$ it is clear that there exists a $\displaystyle N$ such that $\displaystyle N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon$.Now based on the above definition we may say that as soon as $\displaystyle \lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon$ which says exactly that not only does $\displaystyle a_n$ converge, but it converges to $\displaystyle \xi$.

That is how I would do it. Is that acceptable?

You may find this thread helpful http://www.mathhelpforum.com/math-he...-analysis.html - Dec 14th 2008, 01:42 AMJhevon
- Dec 14th 2008, 09:09 AMMathstud28
No I don't $\displaystyle \limsup a_n=\sup\left(S\right)$ where $\displaystyle S$ is the set of all subsequential limits. The same goes for $\displaystyle \liminf a_n=\inf\left(S\right)$

Quote:

secondly, your proof suggests that the sequence is constant, which of course, is not true for convergent sequences in general.

- Dec 14th 2008, 01:39 PMJhevon