- #1
omyojj
- 37
- 0
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
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