Yes, it's true that the set of reals with irrationality measure > 2 has measure 0. See, for example, the end of the nice article by Frits Beukers about pi: www.staff.science.uu.nl/~beuke106/Pi-artikel.ps . Victor The irrationality measure of a real number x, is the supremum of all e such that | x - m/n| < 1/n^e has an infinite number of solutions in rationals m/n. So, in particular, Liouville numbers have infinite irrationality measure. On Sat, Dec 28, 2013 at 12:49 PM, Warren D Smith <warren.wds@gmail.com>wrote:
As VM said, the only way the limit can fail to be 1 is if pi has occasional incredibly close rational approximations. If the role of pi were replaced by some evilly constructed "Liouville-like" transcendental number like SUM 10^(-2^(2^N)) then the limit would fail to exist.
However, I think one can show that if pi is replaced by X then for almost all reals X, the limit exists and is 1. Also, for X any algebraic irrational, ditto. I.e. the set of "bad" real numbers X which have got an infinity of unbelievably close rational approximations, is a set of measure zero. To prove this, consider the standard continued fraction expansion of X and use the Wirsing/Khinchin probability distribution argue to that it is probability 0 that it has an infinite subsequence of large-enough partial quotients. For X algebraic you can use Roth theorem http://mathworld.wolfram.com/RothsTheorem.html
Returning to pi, http://mathworld.wolfram.com/IrrationalityMeasure.html indicates pi has a most a finite set of rational approxs p/q with |pi-p/q| < q^(-7.7). Therefore, in view of limit(N-->oo) N^(7.7/N)=1, the limit exists and is 1.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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