quasi
Posts:
12,067
Registered:
7/15/05


Re: Sequence limit
Posted:
Nov 9, 2013 4:55 AM


Bart Goddard wrote: >Paul wrote in >>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."
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 as 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.
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}.
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

