Am 14.11.2013 06:06, schrieb Virgil: > In article <email@example.com>, > Timothy Murphy <firstname.lastname@example.org> 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
lim_n ->oo e^(-n^(1+eps))^(1/n) = 0
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).