```Date: Oct 3, 2013 4:56 PM
Author: quasi
Subject: Re: Sequence limit

quasi wrote:>quasi wrote:>>konyberg 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. >>>> >>>> Calculus techniques imply the answer is 1.>>>> But the same techniques imply the answer is 1>>>> if n is changed to x, a real variable, and that>>>> is not the case, since sin x =0 infinitely often.>>>> >>>> Anyone wrestled with the subtlies of this problem?>>>> >>>> E.g., can you construct a subsequence n_k such>>>> that sin (n_k) goes to zero so fast that the>>>> exponent can't pull it up to 1?>>>>>>In general a^0 = 1. lim (n goes inf) 1/n = 0. Then the value>>>of sin(n) doesn't change that a^0 = 1.>>>>Your logic is flawed.>>>>Let f(n) = 1/(2^n).>>>>Then f(n)^(1/n) = 1/2 for all nonzero values of n, hence the>>limit, as n approaches infinity, of f(n)^(1/n) is 1/2, not 1.>>>>Are there infinitely many positive integers n such that>>>>   |sin(n)|^(1/n) < 1/(2^n)>>I meant:>>Are there infinitely many positive integers n such that>>   |sin(n)| < 1/(2^n)>>>??>>>>If so, then the limit of the sequence>>>>   |sin(n)|^(1/n), n = 1,2,3, ...>>>>does not exist. In particular, it would not be equal to 1.>>>>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.I think a comparison of power series might a good way toattack the problem.For that approach, I would revise the question as follows:Does there exist a positive real number c such that theinequality   |sin(n)| < exp(-c*n)holds for infinitely many positive integers n?Alternatively, we could ask this question instead:Does there exist a positive real number c such that theinequality   sin^2(n) < exp(-c*n)holds for infinitely many positive integers n?quasi
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