Well, how about 0 1 k k+1 2k 2k+1 ... Then (for x < 1) the alternating sum is just (1-x)/(1-x^k) which is 1/(1+x+x^2+ ... x^(k-1)) As x->1, this -> 1/k. --ms Gareth McCaughan wrote:
On Wednesday 18 May 2005 19:33, dasimov@earthlink.net wrote:
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1.
So. Call an infinite set of non-negative integers "good" if the alternating power series F constructed from it in the same way has lim { x -> 1 from below } F(x) existing. We've established that the powers of 2 aren't good. It's well known that the full set of positive integers is good; the limit is 1/2.
Which sets are good? Can the limit exist but not be 1/2?