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

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