anonymous
  • anonymous
Can I figure out if this sequence is converging with the squeeze theorem? Lim n-->inf. (ln n)^2 /n ?
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
anonymous
  • anonymous
ln(n)^2/n * sorry
anonymous
  • anonymous
I believe he means (ln(n))^2

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anonymous
  • anonymous
Not ln(n)^2
Kainui
  • Kainui
Yeah, ln^2(n) is >1 so you can't use that.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))
anonymous
  • anonymous
Oh yeah I see it it could be greater than one, stupid question :/
Kainui
  • Kainui
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.
anonymous
  • anonymous
Yeah, just noticed it, anyway thanks, basic math issues I guess ^^
Kainui
  • Kainui
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
anonymous
  • anonymous
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
Kainui
  • Kainui
Haha there's no "s" in L'Hopital.
anonymous
  • anonymous
Yikes. Anyway thanks hehe!!
anonymous
  • anonymous
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
Kainui
  • Kainui
Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.
anonymous
  • anonymous
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.
Kainui
  • Kainui
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 ?

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