- #1
transgalactic
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here is the full question and the last step to which i understand
i don't know how to go further??
http://img147.imageshack.us/img147/3077/14993551of0.gif
i got this suggestion from hallsofivy :
You know that for every n, [itex]a_n- L- \epsilon< a_{n+1}< an+ L+ \epsilon[/itex] so [itex]a_n- k(L+ \epsilon)< a_{n+k}< a_n+ k(L+ \epsilon)[/itex] is certainly true for k= 1. Now suppose [itex]a_n- k(L+ \epsilon)< a_n< a_{n+1}+ k(L+ \epsilon)[/itex] is true for some specific k and all n. Then [itex]a_n- (k+1)(L+ \epsilon)= [a_n- (L+ \epsilon)]- k(L+\epsilon)< a_{n+1}-k(L+\epsilon)[/itex] and now use [itex]a_n- k(L+ \epsilon)< a_{n+1}[/itex] with n+1 instead of n- which you can do because it is true for all n.
but i can't see in it my base (k) expression and using it to prove the (k+1) expression
i don't know how to apply it the step i got stuck
??
i don't know how to go further??
http://img147.imageshack.us/img147/3077/14993551of0.gif
i got this suggestion from hallsofivy :
You know that for every n, [itex]a_n- L- \epsilon< a_{n+1}< an+ L+ \epsilon[/itex] so [itex]a_n- k(L+ \epsilon)< a_{n+k}< a_n+ k(L+ \epsilon)[/itex] is certainly true for k= 1. Now suppose [itex]a_n- k(L+ \epsilon)< a_n< a_{n+1}+ k(L+ \epsilon)[/itex] is true for some specific k and all n. Then [itex]a_n- (k+1)(L+ \epsilon)= [a_n- (L+ \epsilon)]- k(L+\epsilon)< a_{n+1}-k(L+\epsilon)[/itex] and now use [itex]a_n- k(L+ \epsilon)< a_{n+1}[/itex] with n+1 instead of n- which you can do because it is true for all n.
but i can't see in it my base (k) expression and using it to prove the (k+1) expression
i don't know how to apply it the step i got stuck
??
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