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December 2nd, 2008, 08:45 AM
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| | 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? | 
December 2nd, 2008, 10:55 AM
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| | Hi, Quote:
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  is  ... | 
December 2nd, 2008, 11:09 AM
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| | If  has no maximal term then ![\left( {\exists N_1 } \right)\left[ 0<{A_1 < A_{N_1 } } \right] \left( {\exists N_1 } \right)\left[ 0<{A_1 < A_{N_1 } } \right]](http://www.mathhelpforum.com/math-help/latex2/img/e5cf5abed40e72fa201ee1097f09c2f4-1.gif) .
Moreover, ![\left( {K > 1} \right)\left( {\exists N_k } \right)\left[ {A_{N_{K - 1} } < A_{N_k } } \right] \left( {K > 1} \right)\left( {\exists N_k } \right)\left[ {A_{N_{K - 1} } < A_{N_k } } \right]](http://www.mathhelpforum.com/math-help/latex2/img/51afa0d2bbb25799a605ec025592486a-1.gif) .
That simply means that the sub sequence  cannot converge to  .
That is a contradiction. | | Thread Tools | | | | Display Modes | Linear Mode |
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