Question

Can I figure out if this sequence is converging with the squeeze theorem?

Lim n-->inf. (ln n)^2 /n ?

Answer 1

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

Answer 2

ln(n)^2/n * sorry

Answer 3

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

Answer 4

Not ln(n)^2

Answer 5

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

Answer 6

oh ok

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

Yikes. Anyway thanks hehe!!

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

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 ?