```Date: Oct 3, 2013 4:33 PM
Author: quasi
Subject: Re: Sequence limit

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?quasi
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