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

 Messages: [ Previous | Next ]
 Roland Franzius Posts: 586 Registered: 12/7/04
Re: Sequence limit
Posted: Oct 4, 2013 4:31 AM

Am 04.10.2013 00:20, schrieb quasi:
> konyberg wrote:
>> quasi wrote:
>>> quasi wrote:
>>>> konyberg wrote:
>>>>> Bart Goddard wrote:
>>>>>>
>>>>>> What is lim_{n -> oo} |sin n|^(1/n)
>>>>>>
>>>>>> where n runs through the positive integers.
>>>>>>
>>>>>> Calculus techniques imply the answer is 1.
>>>>>> But the same techniques imply the answer is 1
>>>>>> if n is changed to x, a real variable, and that
>>>>>> is not the case, since sin x =0 infinitely often.
>>>>>>
>>>>>> Anyone wrestled with the subtlies of this problem?
>>>>>>
>>>>>> E.g., can you construct a subsequence n_k such
>>>>>>
>>>>>> that sin (n_k) goes to zero so fast that the
>>>>>> exponent can't pull it up to 1?

>>>>>
>>>>> In general a^0 = 1. lim (n goes inf) 1/n = 0. Then the value
>>>>> of sin(n) doesn't change that a^0 = 1.

>>>>
>>>>
>>>> Let f(n) = 1/(2^n).
>>>>
>>>> Then f(n)^(1/n) = 1/2 for all nonzero values of n,
>>>> hence the limit, as n approaches infinity, of
>>>> f(n)^(1/n) is 1/2, not 1.

>>>
>>> Are there infinitely many positive integers n such that
>>>
>>> |sin(n)| < 1/(2^n)
>>>

>>>> ??
>>>>
>>>> If so, then the limit of the sequence
>>>>
>>>> |sin(n)|^(1/n), n = 1,2,3, ...
>>>>
>>>> does not exist. In particular, it would not be equal to 1.
>>>>
>>>> In fact, the original question can be recast as:
>>>>
>>>> Does there exist a real number c with 0 < c < 1 such that
>>>>
>>>> the inequality
>>>>
>>>> |sin(n)| < c^n
>>>>
>>>> holds for infinitely many positive integers n?

>>
>> Yes I was a bit hasty here. But sin(n) is limited from -1 to +1

>
> Sure it is.
>
> For positive integers n, the function
>
> f(n) = 1/(2^n)
>
> satisfies 0 < f(n) <= 1/2
>

>> and the limit of 1/n is 0. I would think that the limit is 1.
>
> I agree that in this case, the limit is probably equal to 1,
>
> But in general, if f,g are functions such that, for all
> positive integers n,
>
> (1) 0 < |f(n)| <= 1
> (2) 0 < g(n)
> (3) g(n) --> 0 as n --> oo
>
> the question as to whether or not the limit, as n --> oo, of
>
> f(n)^g(n)
>
> exists, and if so, to what value, cannot be answered without
>
> In particular, for this question, it doesn't matter in the
> least whether the expression 0^0 is regarded as either
>
> undefined
> equal to 1
> equal to 0
> equal to 1/2 (hey, split the difference)
> equal to some other constant
>

>> The debate will still be what 0^0 is equal to :)
>
> Which has no relevance to the OP's question.

Take any sequence of rational approximations of pi
r_i = m_k/n_k -> pi eg the continued fraction for atan 1.

Since pi is not rational, there is no upper limit on n_k. The continued
fractions are the best rational approximations, the limit is never more
than 1/n_k away.

Take a subsequence with denominator n_(k_j) of more than exponential
growth eg e^(n_k_j^2).

Then abs(sin n_(k_j)) = abs(sin ( m_(k_j) pi+ eps_j)) = abs(sin eps_j) <
1/2 eps_j

shows that the left limit point 0 stays to be an accumulation point of
the integer subsequence and therefore of the orginal sequence.

This analysis yields the conjecture that

lim inf (abs (sin (n)))^(1/n) = 0

--

Roland Franzius

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