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Homework Help: Something i don't understand

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    ln(n) < n1/4 for n > N0

    i missed this lecture, so i cant understand what N0 exactly mean

    can anyone tell me what it is?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 17, 2010 #2
    or, is it natural number that start from 0?
     
  4. Apr 17, 2010 #3
    Is that easy for you?
     
  5. Apr 17, 2010 #4
    easy? i don't understand what you are trying to say
     
  6. Apr 17, 2010 #5

    lanedance

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    any context to the statement?

    why not see if you can test if/under what constraints the statement is true...
     
  7. Apr 17, 2010 #6

    D H

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    Better stated,

    [tex]\exists N\in \mathbb N \,:\,\ln n < n^{1/4}\,\forall\,n>N_0[/tex]

    In English, there is some positive integer N0 such that [itex]\ln n < n^{1/4}[/itex] for all n>N0.

    In other words N0 is some integer; you probably need to find it.
     
  8. Apr 17, 2010 #7
    im actually finding whether [tex]\sum[/tex] [tex]\frac{(ln n)^3}{n^3}[/tex] converges or diverges.. and im using comparison test

    ln n < n1/4

    (ln n)3 < n3/4

    => [tex]\frac{(ln n)^3}{n^3}[/tex] < [tex]\frac{1}{n^(9/4)}[/tex]

    since [tex]\sum[/tex][tex]\frac{1}{n^(9/4)}[/tex] converges

    by comparison test [tex]\sum[/tex][tex]\frac{(ln n)^3}{n^3}[/tex] also converges
    -----------------------------------------------------------------------

    ln n < n1/4 do satisfy for all integer n > 0

    so as

    ln n < n1/k , k>0 do also satisfy for all integer n > 0.

    because,

    lim (n->infinity) [tex]\frac{ln n}{n^(1/k)}[/tex] for k>0

    = lim (n->infinity) [tex]\frac{k}{n^1/k}[/tex] (using Lhospital rule)

    = 0 < 1

    => [tex]\frac{ln n}{n^(1/k)}[/tex] < 1 for all k>0

    and it satisfy for all values of n>0

    but if the above state that n > N0
    its not integer start with 0 right? because n=0 is undefine

    or did i do any mistakes?
     
  9. Apr 17, 2010 #8

    Cyosis

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    Try n=5.
     
  10. Apr 17, 2010 #9
    waaaaaaaa, so its wrong then, where did i make mistake?
     
  11. Apr 17, 2010 #10

    Cyosis

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    My point was to give you a simple self check so that you could see that your starting assumption of ln n<n^(1/4) for all n>0 is wrong. So everything that is based on that is wrong as well. As for the rest it is still unclear what you want to do. You have written down some inequality in your original post, but now it appears that you want to figure out if a series converges or not. You would do well by posting the full problem first, then showing your work. In other words please use the template.
     
    Last edited: Apr 17, 2010
  12. Apr 17, 2010 #11

    D H

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    You mistakenly assumed N0 is zero. It's not.
     
  13. Apr 17, 2010 #12
    yea, this is all wrong. owho

    but this

    i really believe this is what i saw on lecture.

    maybe N0 is a different thing, is it?
     
  14. Apr 17, 2010 #13
    yea, this is all wrong. owho

    but this

    i really believe this is what i saw on lecture.

    maybe N0 is a different thing, is it?
     
  15. Apr 17, 2010 #14

    D H

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    Have you read my posts, annoymage?
     
  16. Apr 17, 2010 #15
    so what is N0 for this statement to be true?

    ln n < n1/4 , for all n>N0

    or my lecturer did mistakes with this statement?
     
  17. Apr 17, 2010 #16

    D H

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    No, your lecturer did not make a mistake. As far as the value of N0, you need to find it.
     
  18. Apr 17, 2010 #17
    yeaa, this one,
    and i really need to read the definition to prove that ln n < n^1/4 , n < N thing
    k, thankyou :D
     
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