7
$\begingroup$

Does $\lim_{n\to\infty}\left|\sin n\right|^\frac1n$ where $n\in\mathbb Z^+$ exist?

How can I determine this using freshman calculus?

$\endgroup$
1
  • $\begingroup$ this reslut is 1 $\endgroup$
    – math110
    Commented Aug 12, 2013 at 1:56

1 Answer 1

2
$\begingroup$

solution: let $m>1$ and $|n-m\pi|<\dfrac{\pi}{2},m<n$,use this $$|n-m\pi|>\dfrac{1}{m^{41}}$$ then $$1>|\sin{n}|=|\sin{(n-m\pi)}|>\dfrac{2}{\pi}|n-m\pi|>\dfrac{2}{\pi n^{41}}$$

where use $$\sin{x}>\dfrac{2}{\pi}x,0<x<\dfrac{\pi}{2}$$

and use $$\lim_{n\to\infty}n^{\frac{1}{n}}=1,\lim_{n\to\infty}a^{\frac{1}{n}}=1(a>0)$$ then we have $$\lim_{n\to\infty}|\sin{n}|^{\frac{1}{n}}=1$$

$\endgroup$
4
  • $\begingroup$ Where did the 41 come from? $\endgroup$
    – Dan
    Commented Aug 12, 2013 at 2:29
  • $\begingroup$ can see" K.Mathler,On the approximation of $\pi$,Indag,Math.,15(1953),30-42 $\endgroup$
    – math110
    Commented Aug 12, 2013 at 2:37
  • $\begingroup$ Better values than 41 is also available, but does not affect proof of this problem anyway. $\endgroup$ Commented Aug 12, 2013 at 2:44
  • $\begingroup$ It looks like there's a typo in the last expression in the second line above: "$n^{41}$" should be "$m^{41}$". $\endgroup$
    – Glenn
    Commented Aug 12, 2013 at 17:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .