Re: limit of a special succession

Thanks!
EVE
"Gerry Myerson" <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> ha scritto nel messaggio
news:gerry-014560.10134017012008@xxxxxxxxxxxxxxxxxxxxx
In article <478dfda5$0$37206$4fafbaef@xxxxxxxxxxxxxxxxxxx>,
"Evelyn 83" <isnardiisnardi@xxxxxxxxxx> wrote:

Do you know the limit of

a_n=|sin n| n^a

for a in R?

The problem is that for every epsilon>0 |sin n|<epsilon inifinte times
in
a_n.

If a < 0 the limit is zero (obviously).

If a = 0 the limit doesn't exist (obviously).

So suppose a > 0.

The question is, how close does sin n come to zero
for natural n? Which is the same as asking, how close
does n come to q pi for natural q? Which is the same
as asking, how close does q / n come to 1 / pi?

By Dirichlet, there are infinitely many n, q such that
| (q / n) - (1 / pi) | < 1 / n^2. For these n,
| q pi - n | < pi / n,
so |sin n| is less than C / n for some constant C.

Thus, if a < 1 then there's no limit, since there's
a subsequence going to zero, and a subsequence
going to infinity.

Now there are some much more difficult bounds known
in the other direction, something like
| (q / n) - (1 / pi) | > C / n^(42)
(but don't trust my memory here - go look for it).
Such bounds imply that if a is large enough then a_n goes
to infinity.

Then there will be a range in between, where presumably
the answer is unknown.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)


.

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