Date: Oct 3, 2013 4:56 PM
Author: quasi
Subject: Re: Sequence limit

quasi wrote:
>quasi wrote:
>>konyberg wrote:
>>>Bart Goddard wrote:
>>>>
>>>> This question from a colleague:
>>>>
>>>> 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.

>>
>>Your logic is flawed.
>>
>>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/n) < 1/(2^n)

>
>I meant:
>
>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?


I suspect the answer is no.

I think a comparison of power series might a good way to
attack the problem.

For that approach, I would revise the question as follows:

Does there exist a positive real number c such that the
inequality

|sin(n)| < exp(-c*n)

holds for infinitely many positive integers n?

Alternatively, we could ask this question instead:

Does there exist a positive real number c such that the
inequality

sin^2(n) < exp(-c*n)

holds for infinitely many positive integers n?

quasi