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