# Thread: limit of ln|sin(n)|/n, n->infinity

1. ## limit of ln|sin(n)|/n, n->infinity

prove: $\lim_{n}\cfrac {ln|sin(n)|} {n} = 0 , n\rightarrow \infty$
or show it is not true.
thanks

2. ## Re: limit of ln|sin(n)|/n, n->infinity

I think the limit does not exist but im not sure.
Try with Lhopital rule , you will get limn->inf from cot(n) , and that does not exist as the function is oscilating to infinity without aproaching any specific value.
Sry for bad english

3. ## Re: limit of ln|sin(n)|/n, n->infinity

Since \displaystyle \begin{align*} \sin{(x)} \end{align*} does not have an infinite limit, then I don't see how \displaystyle \begin{align*} \ln{ \left| \sin{(x)} \right| } \end{align*} could possibly have one. So I would expect that the limit does not exist. Thus L'Hospital's Rule should not be used...

4. ## Re: limit of ln|sin(n)|/n, n->infinity

$0 \ge \lim_{n \to \infty} a_{n} \geq \limsup_{n \to \infty} a_{n} = 0$

5. ## Re: limit of ln|sin(n)|/n, n->infinity

Originally Posted by Cartesius24

$0 \geq \displaystyle{\lim_{n \to \infty}} a_{n} \geq \displaystyle{\limsup_{n \to \infty}}~ a_{n} = 0$
corrected latex for our interpreter

6. ## Re: limit of ln|sin(n)|/n, n->infinity

Originally Posted by Cartesius24
$0 \ge \lim_{n \to \infty} a_{n} \geq \limsup_{n \to \infty} a_{n} = 0$
May I ask what you got for the supremum of this sequence?

7. ## Re: limit of ln|sin(n)|/n, n->infinity

I would try to show that for any fixed $\varepsilon>0$ there are infinitely many n's such that $\sin x>1-\varepsilon$ and infinitely many n's such that $|\sin x|<\varepsilon$.

The basic idea is to use Dirichlet principle.

Similar approach was used here: Sine function dense in [−1,1] at MSE.

8. ## Re: limit of ln|sin(n)|/n, n->infinity

Originally Posted by Prove It
May I ask what you got for the supremum of this sequence?
Doen't matter . It is 1. More important is this is finite.