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MarcLeclair

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

  • 11 months ago
  • 11 months ago

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  1. MarcLeclair
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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

    • 11 months ago
  2. MarcLeclair
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    ln(n)^2/n * sorry

    • 11 months ago
  3. genius12
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    I believe he means (ln(n))^2

    • 11 months ago
  4. genius12
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    Not ln(n)^2

    • 11 months ago
  5. Kainui
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    Yeah, ln^2(n) is >1 so you can't use that.

    • 11 months ago
  6. genius12
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    oh ok

    • 11 months ago
  7. MarcLeclair
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    Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))

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

    • 11 months ago
  9. Kainui
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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.

    • 11 months ago
  10. MarcLeclair
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    Yeah, just noticed it, anyway thanks, basic math issues I guess ^^

    • 11 months ago
  11. Kainui
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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.

    • 11 months ago
  12. MarcLeclair
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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

    • 11 months ago
  13. Kainui
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    Haha there's no "s" in L'Hopital.

    • 11 months ago
  14. MarcLeclair
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    Yikes. Anyway thanks hehe!!

    • 11 months ago
  15. genius12
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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

    • 11 months ago
  16. Kainui
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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.

    • 11 months ago
  17. genius12
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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.

    • 11 months ago
  18. Kainui
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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 ?

    • 11 months ago
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