
Re: Sequence limit
Posted:
Oct 4, 2013 4:31 AM


Am 04.10.2013 00:20, schrieb quasi: > konyberg wrote: >> quasi wrote: >>> quasi wrote: >>>> konyberg wrote: >>>>> Bart Goddard wrote: >>>>>> >>>>>> 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/(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? >> >> Yes I was a bit hasty here. But sin(n) is limited from 1 to +1 >> (your function isn't limited), > > Sure it is. > > For positive integers n, the function > > f(n) = 1/(2^n) > > satisfies 0 < f(n) <= 1/2 > >> and the limit of 1/n is 0. I would think that the limit is 1. > > I agree that in this case, the limit is probably equal to 1, > > But in general, if f,g are functions such that, for all > positive integers n, > > (1) 0 < f(n) <= 1 > (2) 0 < g(n) > (3) g(n) > 0 as n > oo > > the question as to whether or not the limit, as n > oo, of > > f(n)^g(n) > > exists, and if so, to what value, cannot be answered without > more information about the functions f,g. > > In particular, for this question, it doesn't matter in the > least whether the expression 0^0 is regarded as either > > undefined > equal to 1 > equal to 0 > equal to 1/2 (hey, split the difference) > equal to some other constant > >> The debate will still be what 0^0 is equal to :) > > Which has no relevance to the OP's question.
Take any sequence of rational approximations of pi r_i = m_k/n_k > pi eg the continued fraction for atan 1.
Since pi is not rational, there is no upper limit on n_k. The continued fractions are the best rational approximations, the limit is never more than 1/n_k away.
Take a subsequence with denominator n_(k_j) of more than exponential growth eg e^(n_k_j^2).
Then abs(sin n_(k_j)) = abs(sin ( m_(k_j) pi+ eps_j)) = abs(sin eps_j) < 1/2 eps_j
shows that the left limit point 0 stays to be an accumulation point of the integer subsequence and therefore of the orginal sequence.
This analysis yields the conjecture that
lim inf (abs (sin (n)))^(1/n) = 0

Roland Franzius

