MAX prove question..

transgalactic · #1 $Old$ 12-02-2008, 08:45 AM

An is a bounded sequence which converges to 0.
An>0 for every n.
prove that in this sequence we have a maximal member
??

again its obvious because if a sequence is descending and it converges from a positive number
to 0.
so 0 is the larger lower bound
and our first member must be the larger

how to transform it to math?

flyingsquirrel · #2 $Old$ 12-02-2008, 10:55 AM

Hi,

Quote:

Originally Posted by transgalactic $View Post$

and our first member must be the larger

A sequence which is convergent and bounded can be non decreasing. For example the maximal member of $(A_n)=\left(\mathrm{e}^{-(n-10)^2}\right)_{n\in\mathbb{N}}$ is $A_{10}=\mathrm{e}^{-(10-10)^2}=1>A_0$ ...

Plato · #3 $Old$ 12-02-2008, 11:09 AM

If $A_n$ has no maximal term then $\left( {\exists N_1 } \right)\left[ 0<{A_1 < A_{N_1 } } \right]$ .
Moreover, $\left( {K > 1} \right)\left( {\exists N_k } \right)\left[ {A_{N_{K - 1} } < A_{N_k } } \right]$ .
That simply means that the sub sequence $\left(A_{N_n}\right)$ cannot converge to $0$ .
That is a contradiction.