Does there exist $\alpha>0$ such that $\lim\limits_{n \to \infty} |n^\alpha \sin n| = \infty$ ?
Asked
Active
Viewed 95 times
1
-
1What are your thoughts? – Did Feb 16 '14 at 17:00
-
It is considered good practice to say what you have tried doing so far, and this will let people help you and give relevant answers. – Joe Tait Feb 16 '14 at 17:11
1 Answers
0
$|n^\alpha sinn|=n^\alpha |sinn|$. Since $lim|sinn|$ has many closure points, (the set of closure points is the segment [0,1]), the sequence $|n^\alpha sinn|$ diverges too. But it does not tend to infinity! It has many closure points. So there is no such $\alpha$ to fill in that equation.
Emo
- 3,446
-
-
Oops, you're right...hehe. Anyway, in the first line I think you mean that the sequence $;{\sin n};$ has many closure points. This though must be proven, and perhaps the OP's doubt is this. – DonAntonio Feb 16 '14 at 17:19