quasi
Posts:
12,042
Registered:
7/15/05


Re: Sequence limit
Posted:
Nov 9, 2013 5:27 AM


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) > > > 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

