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)