Date: Oct 3, 2013 5:42 PM
Author: quasi
Subject: Re: Sequence limit
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.

>> >

>> >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/(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

>(your function isn't limited),

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

more information about the functions f,g.

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.

quasi