1. ## MAX prove question..

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?

2. Hi,
Originally Posted by transgalactic
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 $\displaystyle (A_n)=\left(\mathrm{e}^{-(n-10)^2}\right)_{n\in\mathbb{N}}$ is $\displaystyle A_{10}=\mathrm{e}^{-(10-10)^2}=1>A_0$...

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