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

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Discussion Overview

The discussion revolves around the limit of the sequence \( a_n = | \sin n |^{1/n} \). Participants explore whether the limit exists and the values of the limit superior (limsup) and limit inferior (liminf) of the sequence. The conversation includes theoretical considerations, mathematical reasoning, and hints for proving claims related to the behavior of the sequence as \( n \) approaches infinity.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the limit of the sequence does not exist and questions whether \( \limsup a_n = 1 \) and \( \liminf a_n = 0 \) can be shown.
  • Another participant suggests graphing the function \( f(x) = | \sin(x) |^{1/x} \) and considering the sequence \( \log(a_n) \) for simplification, mentioning the use of l'Hôpital's Rule.
  • A different participant agrees with the graphical observation but raises concerns about the existence of integers \( n \) close to multiples of \( \pi \) that could affect the limit behavior.
  • One participant asserts that the limsup is indeed 1 and questions the liminf, suggesting the use of Diophantine approximation to analyze how often \( | \sin(n) | \) approaches 0.
  • Another participant expresses uncertainty about whether the liminf is 0 or 1, discussing the bounded nature of the sequence and proposing a method involving finite sets of integers.
  • A participant emphasizes that \( a_n \) is never 0 for positive integers \( n \) and cautions about the domain of the graph being restricted to integers.
  • One participant disputes the claim that the liminf could be 0, implying that they have provided a method for computing the limit.
  • Another participant reiterates the evaluation of \( \lim \log(a_n) \) and notes the necessity of having an indeterminate form to apply l'Hôpital's Rule.

Areas of Agreement / Disagreement

Participants express differing views on the existence of the limit and the values of limsup and liminf, with no consensus reached on whether the liminf is 0 or 1. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants note the bounded nature of the sequence and the implications of the equidistribution theorem, but there are unresolved mathematical steps and assumptions regarding the behavior of \( | \sin(n) | \) as \( n \) varies.

omyojj
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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
 
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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.
 
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...
 
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).
 
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..?
 
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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.
 
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No, the liminf isn't 0. I essentially told you how to compute the limit, did you try that?
 
omyojj said:
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...

You can only apply L'Hopital's rule if you have in indeterminate form.
 

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