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MarcLeclair Group Title

Can I figure out if this sequence is converging with the squeeze theorem? Lim n-->inf. (ln n)^2 /n ?

  • one year ago
  • one year ago

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  1. MarcLeclair Group Title
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    I did this \[0 \le \ln(n)^{2} \le 1/n \] solving for the limit and both side gives 0 therefore it must give 0? :/ or am I doing something wrong

    • one year ago
  2. MarcLeclair Group Title
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    ln(n)^2/n * sorry

    • one year ago
  3. genius12 Group Title
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    I believe he means (ln(n))^2

    • one year ago
  4. genius12 Group Title
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    Not ln(n)^2

    • one year ago
  5. Kainui Group Title
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    Yeah, ln^2(n) is >1 so you can't use that.

    • one year ago
  6. genius12 Group Title
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    oh ok

    • one year ago
  7. MarcLeclair Group Title
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    Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))

    • one year ago
  8. MarcLeclair Group Title
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    Oh yeah I see it it could be greater than one, stupid question :/

    • one year ago
  9. Kainui Group Title
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    It does. ln^2(n) is the same as (ln(n))^2. Putting the squared on the function means that. It's so you don't confuse things like sin^2(x) and sin(x^2). But anyways, what I'm saying Marc is that you can't say: ln^2(n)/n≤1/n because it isn't true.

    • one year ago
  10. MarcLeclair Group Title
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    Yeah, just noticed it, anyway thanks, basic math issues I guess ^^

    • one year ago
  11. Kainui Group Title
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    Yeah lol ok so your left hand side should also be something better than 0 I think, that way you're actually squeezing. ln(n)/n<ln^2(n)/n for the right side I'm still thinking haha.

    • one year ago
  12. MarcLeclair Group Title
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    Well I could use l'hospital rule, I just use it so often and my answer key doesn't which gives me wrong answers so I'm afraid to use it now x.x

    • one year ago
  13. Kainui Group Title
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    Haha there's no "s" in L'Hopital.

    • one year ago
  14. MarcLeclair Group Title
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    Yikes. Anyway thanks hehe!!

    • one year ago
  15. genius12 Group Title
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    L'hopital's rule would be much easier to apply here than the squeeze theorem:\[\lim_{n \rightarrow \infty}\frac{ \ln(n)^2 }{ n }=2\lim_{n \rightarrow \infty}\frac{ \ln(n) }{ n }=2\lim_{n \rightarrow \infty}\frac{ 1 }{ n }=2*0=0\] @MarcLeclair @Kainui

    • one year ago
  16. Kainui Group Title
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    Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.

    • one year ago
  17. genius12 Group Title
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    Well you are supposed to use squeeze theorem which he figured already, but l'hopitals would be easier if we were allowed to use it.

    • one year ago
  18. Kainui Group Title
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    Haha yeah for sure. Actually we can just use that simple idea to squeeze it between these two which are easy to also find the limit of with l'hopital's! haha ln(n)/n≤ln^2(n)/n≤ln^3(n)/n ?

    • one year ago
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