From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math
Subject: Re: Sequence {(sin n)^n}  - Converge?
Date: 9 Nov 2000 19:24:40 GMT
Organization: ITServices, University of British Columbia
Lines: 23
Message-ID: <8uetlo$4t8$1@nntp.itservices.ubc.ca>
References: <20001108205813.12103.00000473@ng-bg1.aol.com>
X-Trace: nntp.itservices.ubc.ca 973797880 5032 137.82.36.62 (9 Nov 2000 19:24:40 GMT)
X-Complaints-To: abuse@interchange.ubc.ca
X-Newsreader: trn 4.0-test72 (19 April 1999)


In article <20001108205813.12103.00000473@ng-bg1.aol.com>,
LMWapner <lmwapner@aol.com> wrote:

>Does the above sequence converge (assuming n is in radians)?

There are infinitely many pairs of positive integers m,n such that 
|pi/2 - n/m| < 1/m^2.  I believe this is true also if m is required to be
 odd (maybe with 1 replaced by some other positive constant).  Then 
|n - m pi/2| < 1/m and |sin n| > 1 - 1/(2 m^2), which in turn implies
|sin n|^n > some positive constant epsilon.  Thus |sin n|^n diverges,
in fact even |sin n|^(n^2) diverges.

On the other hand, by a result of Hata |pi/2 - n/m| > 1/m^8.0161 for
sufficiently large m, so |sin n|^(n^14.0322) converges.  
For almost every real x and any epsilon > 0, 
|pi/(2x) - n/m| > 1/m^(2+epsilon/2) 
for sufficiently large m, implying that |sin(nx)|^(n^(2+epsilon))
converges.   

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2