quasi
Posts:
9,903
Registered:
7/15/05


Re: Sequence limit
Posted:
Oct 3, 2013 4:56 PM


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 to attack the problem.
For that approach, I would revise the question as follows:
Does there exist a positive real number c such that the inequality
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 the inequality
sin^2(n) < exp(c*n)
holds for infinitely many positive integers n?
quasi

