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prove if (a_n)--> infinity then (1/a_n) --> infinity watch

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    (Original post by hiyatt)
    is there a nicy easy fix to this without re doing the whole thing
    The whole proof is about 3 lines! How easy do you want it to be?
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    (Original post by DFranklin)
    The whole proof is about 3 lines! How easy do you want it to be?
    lool you're funny. can i just let e=1/C at the end so it matches.
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    (Original post by DFranklin)
    The thing is, it can be hard to distinguish between inability to put everything in a coherent logical order, and just a reluctance to actually post any mathematics here.

    Or at least, posters frequently post multiple single line "so, should I do this...?" type questions, and you have no idea what their actual proof looks like anymore...
    Very true!
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    (Original post by hiyatt)
    lool you're funny. can i just let e=1/C at the end so it matches.
    No. That's the 3rd time you've asked if you can do this. The answer isn't going to change.
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    (Original post by ftfy)
    Very true!
    That post of mine is looking staggeringly prophetic given the OPs most recent postings...
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    (Original post by DFranklin)
    No. That's the 3rd time you've asked if you can do this. The answer isn't going to change.
    I have said at the begginging to let C>0 so if you do that you see e varies to in the required range though
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    (Original post by hiyatt)
    I have said at the begginging to let C>0 so if you do that you see e varies to in the required range though
    That's not the way around it works. You have to *start* by choosing an arbitrary epsilon > 0. Once you do that, it's fixed - you can't set it to 1/C later.
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    (Original post by DFranklin)
    That's not the way around it works. You have to *start* by choosing an arbitrary epsilon > 0. Once you do that, it's fixed - you can't set it to 1/C later.
    I have to think about this later
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    (Original post by hiyatt)
    I have to think about this later
    You should think in particular about the question I asked you in post 17:
    Now, you want to show |1/a_n| < epsilon (for sufficiently large n). What condition does this put on a_n?
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    (Original post by DFranklin)
    That post of mine is looking staggeringly prophetic given the OPs most recent postings...
    Indeed! :lol:
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    (Original post by DFranklin)
    You should think in particular about the question I asked you in post 17:


    can you do the work to get to C/2 and then say if we choose a e>0 there exists an N such that we have l1/a_nl<e whenever n>N
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    (Original post by hiyatt)
    can you do the work to get to C/2 and then say if we choose a e>0 there exists an N such that we have l1/a_nl<e whenever n>N
    If you think you have a proof, post a proof. It's impossible to judge whether a partially complete fragment would "work", although the fact that you talk about C/2 makes me pretty certain you've done something wrong.

    Again, I ask you to answer the question I asked in post #17.
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    (Original post by DFranklin)
    If you think you have a proof, post a proof. It's impossible to judge whether a partially complete fragment would "work", although the fact that you talk about C/2 makes me pretty certain you've done something wrong.

    Again, I ask you to answer the question I asked in post #17.
    Ok if this is not correct i will answer the question.

    Prove 'If (a_n) --> infinity then (1/a_n) --> 0

    As a_n--> infinity, for every C>0 there exists an N in the natural numbers such that we have

    a_n>C whenever n>N... this implies

    la_nl>C whenever n>N... this implies

    l1/a_nl < 1/C whenever n>N

    If follows that if we choose e>0 there exists an N_1 such that we have

    l1/a_nl < e whenever n>N_1
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    (Original post by hiyatt)
    Ok if this is not correct i will answer the question.

    Prove 'If (a_n) --> infinity then (1/a_n) --> 0

    As a_n--> infinity, for every C>0 there exists an N in the natural numbers such that we have

    a_n>C whenever n>N... this implies

    la_nl>C whenever n>N... this implies

    l1/a_nl < 1/C whenever n>N

    If follows that if we choose e>0 there exists an N_1 such that we have

    l1/a_nl < e whenever n>N_1
    It's still not right. A kind examiner might fill in the gaps, but he shouldn't have to. The line where you says "it follows" is particularly dubious.
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    (Original post by DFranklin)
    It's still not right. A kind examiner might fill in the gaps, but he shouldn't have to. The line where you says "it follows" is particularly dubious.
    can you write what you would say...
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    (Original post by hiyatt)
    can you write what you would say...
    No. In fact, I'm done with this. I can't be bothered repeating myself 5 times on every single thiing because you don't want to believe what I tell you.
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    (Original post by DFranklin)
    No. In fact, I'm done with this. I can't be bothered repeating myself 5 times on every single thiing because you don't want to believe what I tell you.
    Haha no of course i believe you you are much mor eintelligent than me i was just wondering if my version of it had any credibility
 
 
 
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