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Math Help - divergence of {an} - help please

  1. #1
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    divergence of {an} - help please

    Hey people.

    I have a series {an}.
    I know that for every positive R > 0 exists an n so |an| >= R, where n is a natural number.

    Now i have to show that {an} is divergent.


    I dont exacly know how to do this. I tried to make a proof by conflict, saying that {an} was convergent and then choosing an R > epsilon + |a| where a is the limit of {an}. Then i thought i might be able to end up with something of the following:
    For every R > 0 there is an n so |an| >= R. If R = epsilon + |a| -> |an| > epsilon + |a| -> |an| - |a| > epsilon..
    Then i didnt know what to do, because i dont know if i can say that |an|-|a| =< |an - a| and then say that |an - a| >= |an| - |a| > epsilon -> |an - a| > epsilon.

    Then it conflicts with the definition of convergence, cause it states that for any epsilon, there exists an N so |an - a| <= epsilon, which in this case isnt true.

    Can i do this?
    Thanks for the help.
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  2. #2
    Member Abu-Khalil's Avatar
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    Let \epsilon>0. Suppose a_n\to A\in\mathbb{R}, then \exists N:\forall n\geq N,|a_n-A|<\epsilon.

    Let r=\max_{1\leq n\leq N}|a_n| then choose R=\max\left\{r+\epsilon,|A|+\epsilon,\right\}.

    The r+\epsilon is to get a new n and then, for a n\geq N we have

    \epsilon>|a_n-A|\geq |a_n|-|A|\geq R+|A|+\epsilon-|A|=R+\epsilon.\Rightarrow 0>R\rightarrow\leftarrow
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  3. #3
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    Ok. I understand the last part.
    I dont understand why you choose r the way you do?
    Could you explain in more detail?

    Thanks.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Abu-Khalil View Post
    Let \epsilon>0. Suppose a_n\to A\in\mathbb{R}, then \exists N:\forall n\geq N,|a_n-A|<\epsilon.

    Let r=\max_{1\leq n\leq N}|a_n| then choose R=\max\left\{r+\epsilon,|A|+\epsilon,\right\}.

    The r+\epsilon is to get a new n and then, for a n\geq N we have

    \epsilon>|a_n-A|\geq |a_n|-|A|\geq R+|A|+\epsilon-|A|=R+\epsilon.\Rightarrow 0>R\rightarrow\leftarrow
    In your last line, how do you justify  |a_n|\geq R+|A|+\epsilon ?
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  5. #5
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    Quote Originally Posted by MortenDK View Post
    I have a series {an}.
    I know that for every positive R > 0 exists an n so |an| >= R, where n is a natural number.
    Now i have to show that {an} is divergent.
    It seems that you are over-thinking this problem.
    The first test for divergence states:  \sum\limits_{k = 1}^\infty  {a_k } converges only if \left( {a_n } \right) \to 0.
    Is that possible in this case in light of the given?
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  6. #6
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    First of all, doesnt it converge if it has a limit? that limit doesnt have to be 0.
    Second, yes i know its divergent, but i have to give a mathematical proof. Its a hand-in so it has to be pretty formal. I just dont know how to explain it proberly.
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  7. #7
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    Quote Originally Posted by MortenDK View Post
    First of all, doesnt it converge if it has a limit? that limit doesnt have to be 0.
    Second, yes i know its divergent, but i have to give a mathematical proof. Its a hand-in so it has to be pretty formal. I just dont know how to explain it proberly.
    Here is the proof that the series diverges.
    From the given \left( {a_n } \right) \not\to 0 therefore \sum\limits_{k = 1}^\infty  {a_k } diverges.
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  8. #8
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    No. A series doesnt have to go to 0 to be convergent. It only has to have a limit, meaning that when n -> infinite an -> x.
    Convergent series - Wikipedia, the free encyclopedia here it can be seen as it is written in my book. It is stated around the 5th line or something like that.

    Also, what is in my assignment doesnt state that it doesnt have a limit, it only states what i wrote in the beginning of my first post.

    Im pretty sure that Abu-Khalil has the solution, i simply dont understand why he defines r the way he does in the middle.
    Last edited by MortenDK; November 27th 2009 at 05:54 PM. Reason: wrote the wrong name :P
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  9. #9
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by chiph588@ View Post
    In your last line, how do you justify  |a_n|\geq R+|A|+\epsilon ?
    Oh thanks, my mistake. It was |a_n|-|A|\geq R-|A|\geq |A|+\epsilon-|A|=\epsilon \Rightarrow \epsilon >\epsilon.

    I choose that R so I can assure that it will be a new n and no a n<N.

    But, i have a confussion i supossed that a_n=\sum_{k=1}^nx_k and given any R>0 we may find n\in\mathbb{N}:|a_n|>R or it was |x_k|\geq R?
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  10. #10
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    Hmm.. dunno if im just daft or something, but i dont understand it.
    Could you rewrite the whole thing, editing your typo?
    Sorry for the inconvinience.
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