## 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

1. MarcLeclair Group Title

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

2. MarcLeclair Group Title

ln(n)^2/n * sorry

3. genius12 Group Title

I believe he means (ln(n))^2

4. genius12 Group Title

Not ln(n)^2

5. Kainui Group Title

Yeah, ln^2(n) is >1 so you can't use that.

6. genius12 Group Title

oh ok

7. MarcLeclair Group Title

Wait I thought (ln(n))^2 meant (ln(n)) * (ln(n))

8. MarcLeclair Group Title

Oh yeah I see it it could be greater than one, stupid question :/

9. Kainui Group Title

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.

10. MarcLeclair Group Title

Yeah, just noticed it, anyway thanks, basic math issues I guess ^^

11. Kainui Group Title

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.

12. MarcLeclair Group Title

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

13. Kainui Group Title

Haha there's no "s" in L'Hopital.

14. MarcLeclair Group Title

Yikes. Anyway thanks hehe!!

15. genius12 Group Title

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

16. Kainui Group Title

Hmm I agree with you here, but I thought the question was specifically asking to solve this with the squeeze theorem.

17. genius12 Group Title

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.

18. Kainui Group Title

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 ?