quasi
Posts:
11,740
Registered:
7/15/05


Re: Sequence limit
Posted:
Nov 9, 2013 6:29 AM


quasi wrote: >quasi wrote: >>Bart Goddard wrote: >>>Paul wrote: >>>>Bart Goddard wrote: >>>>> Timothy Murphy wrote: >>>>> > >>>>> > It's clear that the upper limit is 1, >> >>Rephrased: >> >> "It's clear that >> >> lim_sup {sin(n)^(1/n)  n in N} = 1." > >In meant: > > "It's clear that lim sup of the sequence > > sin(n)^(1/n), n = 1,2,3, ... > > is equal to 1." > >>I agree  it's obvious, or at least elementary. >> >>The fact that the lim sup equals 1 can be easily shown without >>invoking the stronger, more advanced result that the actual >>limit exists and is equal to 1. >> >>>>> > since n pi mod Z will be distributed evenly in [0,1), >> >>Rephrased: >> >> "The set {(n*Pi) mod 1  n in N} is dense in (0,1)." >> >>>>> > and so will infinitely often be in the range (1/3,2/3). >> >>Yes, true. >> >>>>> I don't think this is true. >> >>I think you misunderstand Tim's claim. >> >>Tim wasn't claiming that the sequence >> >> sin(n)^(1/n), n = 1,2,3, ... >> >>has infinitely many elements in the interval (1/3,2/3). >> >>That would be false since the limit is 1. >> >>Rather, he was claiming that the sequence >> >> (n*Pi) mod 1, n = 1,2,3, ... >> >>has infinitely many elements in the interval (1/3,2/3). >> >>That claim is true, and, as an easy consequence, it follows >>that lim_sup sin(n)^(1/n)  n in N} = 1. > >I meant: > >lim sup of the sequence > > sin(n)^(1/n), n = 1,2,3, ... > >is equal to 1. > >>Thus, the lim sup is easy. >> >>By contrast, the limit is hard. >> >>>>>Sin(n) is probably distributed evenly, >> >>Yes, in the sense that the set >> >> {sin(n) : n in N} >> >>is dense in the interval (0,1). >> >>>>>but raising to (1/n) power is going to crowd things toward 0. >> >>No, it will crowd things toward 1. >> >>>> This sounds ambiguous to me. >>>> >>>> Do you mean that you don't think it's true that the >>>> upper limit is 1. >> >>We're way past that. >> >>In this thread, it's already been established that the limit is >>known to be equal to 1, so of course the lim sup is also 1. >> >>>or do you mean that you don't think it's true that it's clear >>>that the upper limit is 1? >>> >>>Neither. >>> >>>Obviously the thing I don't think is true is the >>>sentence to which I'm responding: >>> >>>That the values of sin n^(1/n) are evenly distributed. >> >>Tim never made that claim. >> >>>You could infer this by my comment about the (1/n) power >>>pushing things toward zero. >>> >>>It has already been proven that the limit is 1. >> >>Right, but Tim is focusing just on the lim sup, using only >>elementary reasoning. >> >>To me, his reasoning seems correct. >> >>But the argument can be made even simpler ... >> >>Let L = lim_sup {sin(n)^(1/n)  n in N}. > >I meant: > >Let L = lim sup of the sequence > > sin(n)^(1/n), n = 1,2,3, ... > >>We wish to show L = 1. >> >>It's clear that sin(n)^(1/n) < 1 for all n in N, hence L <= 1. >> >>lemma: >> >>For any two consecutive terms of the sequence >> >> sin(n), n = 1,2,3, ... >> >>at least one of them exceeds 2/5. >> >>Proof: >> >>Suppose n is a positive integer such that sin(n) <= 2/5. >> >>We will show that sin(n+1) > 2/5. >> >>Firstly, >> >> => sin(n) <= 2/5 >> >> => sin^2(n) <= 4/25 >> >> => cos^2(n) >= 21/25 >> >>Then >> >> sin^2(n + 1) >> >> = (sin(n)cos(1) + cos(n)sin(1))^2 >> >> = sin^2(n)cos^2(1) >> + cos^2(n)sin^2(1) >>  2sin(n)cos(n)sin(1)cos(1)
Of course that last line above should be
+ 2sin(n)cos(n)sin(1)cos(1)
but that correction doesn't affect what follows.
>> > cos^2(n)sin^2(1)  2sin(n)cos(n)sin(1)cos(1) >> >> >= (21/25)sin^2(1)  2(2/5)(1)sin(1)cos(1) >> >> > 4/25 >> >>Hence sin(n+1) > 2/5, as was to be shown. >> >>It follows that >> >> (2/5)^(1/n) < sin(n)^(1/n) < 1 >> >>holds for infinitely many positive integers n. >> >>Since (2/5)^(1/n) > 1 as n > oo, it follows that L = 1.
quasi

