
Re: Sequence limit
Posted:
Oct 4, 2013 11:37 AM


On Thu, 03 Oct 2013 16:48:16 0500, quasi <quasi@null.set> wrote:
>quasi wrote: >>quasi wrote:
>>>>Bart Goddard wrote: >>>>> >>>>> This question from a colleague: >>>>> >>>>> What is lim_{n > oo} sin n^(1/n) >>>>> >>>>> where n runs through the positive integers.
>>>In fact, the original question can be recast as: >>> >>>Does there exist a real number c with 0 < c < 1 such that >>>the inequality >>> >>> sin(n) < c^n >>> >>>holds for infinitely many positive integers n? > >I suspect the answer is no.
Assume that such a c exists and that the inequality holds for infinitely many n..
For such an n there is an integer k for which k pi  n <= pi/2 and
therefore k pi  n <= pi/2 sin n < 2 c^n.
From pi  n/k < 2/k c^n it follows that n/k converges towards pi and that for sufficientlly large n one has n > 2 k, which leads to
pi  n/k < 2/k c^(2k).
Simple analysis shows that for any m > 1 the expression 2/x c^(2x) x^m converges with limit 0 when x > oo, so for sufficiently large n (and k) one has
2/k c^(2k) k^m <1 and pi  n/k < 1/k^m
In other words, pi would have an infinite irrationality measure. However, the MathWorld page about the irrationality measure, among others, mentions a finite upper bound for the irrationality measure of pi, so the initial assumption does not hold.
Leon

