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

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 quasi Posts: 12,042 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.

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}.

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