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Topic: Sequence limit
Replies: 72   Last Post: Nov 26, 2013 12:07 AM

 Messages: [ Previous | Next ]
 quasi Posts: 12,067 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.
>
>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.

>
>

>>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

Date Subject Author
10/3/13 Bart Goddard
10/3/13 Karl-Olav Nyberg
10/3/13 quasi
10/3/13 quasi
10/3/13 Karl-Olav Nyberg
10/3/13 quasi
10/4/13 Roland Franzius
10/4/13 quasi
10/5/13 Roland Franzius
10/5/13 quasi
10/26/13 Roland Franzius
10/26/13 karl
10/26/13 Roland Franzius
10/26/13 gnasher729
10/27/13 karl
10/3/13 quasi
10/4/13 Leon Aigret
10/4/13 William Elliot
10/4/13 quasi
10/4/13 William Elliot
10/4/13 quasi
10/4/13 David C. Ullrich
10/4/13 Robin Chapman
10/5/13 Bart Goddard
10/4/13 Bart Goddard
10/4/13 Peter Percival
10/5/13 Virgil
10/4/13 Bart Goddard
10/6/13 David Bernier
10/6/13 Virgil
10/6/13 Bart Goddard
10/7/13 Mohan Pawar
10/7/13 Bart Goddard
10/7/13 gnasher729
10/7/13 Richard Tobin
10/7/13 Robin Chapman
10/7/13 Michael F. Stemper
10/7/13 Michael F. Stemper
10/7/13 David Bernier
10/7/13 fom
10/8/13 Virgil
10/8/13 fom
10/8/13 Virgil
10/8/13 fom
10/4/13 fom
10/4/13 quasi
10/4/13 quasi
10/9/13 Shmuel (Seymour J.) Metz
10/10/13 Bart Goddard
11/5/13 Shmuel (Seymour J.) Metz
11/6/13 Bart Goddard
11/11/13 Shmuel (Seymour J.) Metz
11/12/13 Bart Goddard
11/15/13 Shmuel (Seymour J.) Metz
11/15/13 Bart Goddard
11/6/13 Timothy Murphy
11/8/13 Bart Goddard
11/8/13 Paul
11/8/13 Bart Goddard
11/9/13 Paul
11/9/13 quasi
11/9/13 quasi
11/9/13 quasi
11/13/13 Timothy Murphy
11/13/13 quasi
11/14/13 Timothy Murphy
11/14/13 Virgil
11/14/13 Roland Franzius
11/26/13 Shmuel (Seymour J.) Metz
11/9/13 Roland Franzius
11/9/13 Paul