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Analysislimsup(inf) |sinn|^1/n = ? it`s not easy

  1. Dec 4, 2007 #1
    Hi,all
    My goal here is to show that the limit of a sequence a_n = |sin n|^1/n
    i.e. lim _ |sin n |^1/n does not exist..(it does?)

    is it possible to show that 1. limsup a_n = 1
    2. liminf a_n = 0 (right??I`m not sure..)

    Help me doing this job..
    I think that it is more difficult to prove the latter..

    one thing that obvious is that limsup |sinn|=1, liminf|sinn|=0..; )

    it may be helpful using the equidistribution theorem..
    (the sequence {1,2,3,...} is equidistributed mod 2π, quoted from the Wikipedia limsup)
    then we can argue that for any 0<=a<b<=1 there are infinitely many n
    such that a<|sin n|<b, and then we can proceed by using the def. of the limsup..

    anyway..give me some hint..or solution..T.T
     
  2. jcsd
  3. Dec 5, 2007 #2

    AKG

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    Graph the function f(x) = |sin(x)|1/x. It's got a pretty neat graph, and also you'll see that as x goes to infinity, f(x) does indeed appear to converge to a limit. So to solve your problem, consider the sequence log(an), make a simplifying comparison, and keep l'Hôpital's Rule in mind.
     
  4. Dec 5, 2007 #3
    that seems to be right on the graph..thx..
    f=|sinx|^1/x is close to 1 on the left for sufficiently large x..(but it does have inf. many zeros at integral multiples of Pi)...
    but if there exists n that is in the δ neigborhood of m*Pi(m integer)...
    such that |sinn|^1/n may not exceed 1-ε..
    and the set of numbers satisfying this condition could be infinite....I don`t know..

    If you are right..that it converges to 1..

    hence limsup a_=1 then possible to show that liminf a_n = 1???

    and if the lim exists..lim log(a_n) also exists..

    To evaluate lim_log a_n = lim_1/x * log[|sinx|]

    (x)` = 1 as x->inf.
    but (log[|sinx|])` = |sinx|`/|sinx| -> ??? x-> inf ??
    clearly the limit does not exist...
     
  5. Dec 5, 2007 #4
    Either way the limsup is 1. Who cares if there are a lot of 0s? There are also a lot of times when it will be very close to 1. A more interesting question is whether the liminf is 0 if n is an integer...

    the hint would be to use a Diophantine approximation result to see how close |sin(n)| gets to 0 and how often. try Hurwitz' irrational number theorem (you can google this).
     
  6. Dec 5, 2007 #5
    surely..I wonder whether the liminf a_n =? 0 or 1? or something..
    (If the terms in the sequence are real numbers, the limit superior and limit inferior always exist in the affinely extended real number system..this seq. is bounded since 0<a_n<1 for all n..so 0<=limsup, liminf<=1 )
    to be sure, a_n is a inf. seq. of real numbers..n is integer..


    To prove liminf a_n = 0..How about this way..
    Fix ε, suppose that there are finite set of integers such that |sin n_k| < ε^n_k
    {n_k} = {n_1, n_2, ..., n_N}
    is there any contradiction which leads to the equipartition thm..or sth..?
     
    Last edited: Dec 5, 2007
  7. Dec 5, 2007 #6
    Again, be careful as a_n = |sin n|^(1/n) is never 0 for any positive integer value of n. For positive integers n, |sin n| is strictly between 0 and 1. This seems silly to point out, but better to be safe than sorry: in your graph, your domain should be restricted to the integers, and not reals. Otherwise you'll see infinitely many zeros.
     
    Last edited: Dec 5, 2007
  8. Dec 5, 2007 #7

    AKG

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    No, the liminf isn't 0. I essentially told you how to compute the limit, did you try that?
     
  9. Dec 5, 2007 #8
    You can only apply L'Hopital's rule if you have in indeterminate form.
     
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