Does $\displaystyle \{ |\sin n|^{\frac{1}{n}}\}$ converge?
Remember that
$\displaystyle -1 \leq \sin{n} \leq 1$ for all $\displaystyle n$, that means
$\displaystyle |\sin{n}| \leq 1$ for all $\displaystyle n$.
I assume since this is a sequence that $\displaystyle n$ only takes on postitive integer values, so that would mean $\displaystyle \frac{1}{n}$ is positive and therefore exponentiating would keep the inequality intact.
So $\displaystyle |\sin{n}|^{\frac{1}{n}} < 1^{\frac{1}{n}}$
$\displaystyle |\sin{n}|^{\frac{1}{n}} < 1$.
So yes, the sequence converges.
Is...
$\displaystyle \displaystyle |\sin n|^{\frac{1}{n}} = e^{\frac{\ln |\sin n|}{n}$ (1)
... and because ...
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0 $ (2)
... is ...
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} |\sin n|^{\frac{1}{n}} = 1$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$
is not obvious. there are integer $\displaystyle n's$ that make $\displaystyle |\sin n|$ very close to $\displaystyle 0$ hence $\displaystyle -\ln |\sin n|$ is very big.
Reply to Elim at Post # 5:
Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.
The consideration of elim is quite correct. By definition say that...
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$ (1)
... means that for any $\displaystyle \varepsilon >0$ it exists an integer $\displaystyle N$ so that $\displaystyle \forall n>N$ is...
$\displaystyle \displaystyle \frac{|\ln |\sin n||}{n}< \varepsilon $ (2)
Now if we measure angles in degree [for example] instead of in radians $\displaystyle \forall n|180$ is $\displaystyle \sin n =0$ and the sequence $\displaystyle \displaystyle a_{n} = |\sin n|^{\frac{1}{n}}$ has no limit. If the angles are in radians then $\displaystyle \sin n$ never vanishes and in my opinion the (1) is 'probably true', even if for $\displaystyle \varepsilon$ 'very small' the (2) is satisfied for 'enormous' values of $\displaystyle N$. In any case a rigorous prove has to be supplied and that is not a trivial job ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$