Date: Oct 4, 2013 10:54 AM
Author: David C. Ullrich
Subject: Re: Sequence limit

On Fri, 04 Oct 2013 05:14:54 -0500, quasi <quasi@null.set> wrote:

>William Elliot wrote:
>>Bart Goddard wrote:
>>

>>> What is lim_{n -> oo} |sin n|^(1/n)
>>>
>>> where n runs through the positive integers.

>>
>>lim(n->oo) |sin n|^(1/n)
>> = lim(n->oo) (|sin n|/n)^1/n) * n^(1/n)
>> = lim(n->oo) (|sin n|/n)^(1/n) * lim(n->oo) n^(1/n)
>> = 1 * 1.

>
>Not proved.
>
>Not even close.
>
>You've given no justification for the claim
>
> lim(n->oo) (|sin n|/n)^(1/n) = 1


And worse, it's obvious that proving that is harder
than proving that the original limit is 1, since sin(n)/n
is closer to 0 than sin(n).

>
>quasi