Date: Oct 3, 2013 4:33 PM
Author: quasi
Subject: Re: Sequence limit

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?


quasi