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Math Help - Raabe Test

  1. #1
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    Question Raabe Test

    Show that if lim(an+1/an)>1 there exists a natural number such that lim(an+1/an)=>1+1/n.

    I have been trying this, but I cant seem to get the proof?

    Any Ideas? Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    Show that if lim(an+1/an)>1 there exists a natural number such that lim(an+1/an)=>1+1/n.

    I have been trying this, but I cant seem to get the proof?

    Any Ideas? Thanks
    Forget the limit part. If \alpha>1\implies \alpha-1>0 and by the Archimedean principle there exist some n\in\mathbb{N} such that \frac{1}{n}<\alpha-1\implies \alpha>1+\frac{1}{n}
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  3. #3
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    So how does this link back to the lim(an+1/an)>1.....? I can see what you are using.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    So how does this link back to the lim(an+1/an)>1.....? I can see what you are using.
    I'm not sure what you mean here? Let \alpha=\lim\text{ }\frac{a_{n+1}}{a_n} assuming it exists.
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  5. #5
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    lim(an+1/an)>1 there exists a natural number such that lim (an+1/an)=>1+1/n..........the () are absolute values.....link in the ratio test...i dont know how to write it....sorry...would this change anything
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    lim(an+1/an)>1 there exists a natural number such that lim (an+1/an)=>1+1/n..........the () are absolute values.....link in the ratio test...i dont know how to write it....sorry...would this change anything
    In our example \alpha is any number. You tell me, does it make a difference?
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  7. #7
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    no I dont think so...its just the proof for lim (an+1/an)<1 looked a bit different in my notes, which yields <= 1-a/n for all n >= N....this is for the Raabe test
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    no I dont think so...its just the proof for lim (an+1/an)<1 looked a bit different in my notes, which yields <= 1-a/n for all n >= N....this is for the Raabe test
    Raabe test for series convergence?
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  9. #9
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    yes...the one you use once ur ratio test yields 1...so we try the raabe test for series
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    yes...the one you use once ur ratio test yields 1...so we try the raabe test for series
    Ok. I understand what Raabe's test is. I am just wondering if your problem is not what it seems. I literally interpreted your question as if \lim\text{ }\left|\frac{a_{n+1}}{a_n}\right|>1 then there exists some n'\in\mathbb{N} such that \lim\text{ }\left|\frac{a_{n+1}}{a_n}\right|\geqslant 1+\frac{1}{n'}
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  11. #11
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    yes this is what I am having a little trouble with...proving this...can u help?
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    yes this is what I am having a little trouble with...proving this...can u help?
    That is exactly what I just proved.
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    SO we let L=lim (an+1/an)>1

    then L>1.....L-1>0...the via the Archimedean principle there exist a natural number N such that L-1>1/n........L=1+1/n......done?
    Last edited by hebby; February 10th 2010 at 09:10 PM.
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  14. #14
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    SO we let L=lim (an+1/an)>1

    then L>1.....L-1>0...the via the Archimedean principle there exist a natural number N such that L-1>1/n........L=1+1/n......done?
    Change that last equals sign to a > and yes.
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