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Topic: Sequence limit
Replies: 72   Last Post: Nov 26, 2013 12:07 AM

 Messages: [ Previous | Next ]
 Roland Franzius Posts: 586 Registered: 12/7/04
Re: Sequence limit
Posted: Nov 14, 2013 4:57 AM

Am 14.11.2013 06:06, schrieb Virgil:
> In article <l614vg$r43$1@dont-email.me>,
> Timothy Murphy <gayleard@eircom.net> wrote:
>

>>
>> You seem to be responding to yourself.
>> I did not say that "|sin n|^(1/n) is evenly distributed".
>>

>>> It has already been proven that the limit is 1.
>>
>> I think I missed this proof.
>> What was the essential idea?

>
> Note that for any fixed value, x, between 0 and 1, lim x ^(1/n) = 1
>

But since eg

lim_n->oo 0^(1/n) = 0

and

lim_n ->oo e^(-n^(1+eps))^(1/n) = 0

and

lim_n->oo (q^n)^(1/n) = q

these ideas only give a hint, that a series
(a_n)^(1/n), a_n \in (0,1)
may have any accumulation points in 0,1.

An accumulation point x < 1 exists,
if there exist inifinite subseries a_(n_k) bounded by a series
q^(n_k) with fixed 0 < q < 1,
which means,
provided that for any N > 1 there exists a q(N), 0 <= q(N) < r < 1
such that there are infinitely many k in the subseries n_k > N with
a_(n_k) <= q^(n_k).

--

Roland Franzius

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