Hello! Older references on this problem are: G. H. Hardy, On certain oscillating series, Quart. J. Pure Appl. Math. 38 (1907) 269-288 G. H. Hardy, Divergent Series, Oxford Univ. Press, 1949, p. 77 Steve Finch http://pauillac.inria.fr/algo/bsolve/
From: "" <dasimov@earthlink.net> Reply-To: dasimov@earthlink.net, math-fun <math-fun@mailman.xmission.com> To: "math-fun" <math-fun@mailman.xmission.com> Subject: RE: [math-fun] Re: A puzzle I read about Date: Wed, 18 May 2005 14:33:02 -0400
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.
--Dan ---------------------------------------------------------------------------------
<< Dan Asimov asked:
For positive x<1, consider the alternating sum S(x) = x - x2 + x4 - x8 + x16 - x32 + - ... Does S(x) approach a limit as x approaches 1 from below, and if so what is this limit?
The limit does not exist. Since S(x) = x - S(x^2), the limit would have to be 1/2 if it existed. But I'll show that there are values of x arbitrarily close to 1 for which S(x) > 0.5008.
Rewrite the definition as
S(x) = SUM (x^4^k - x^(2*4^k)) k>=0
For 0 < x < 1, the summands are all positive, so any partial sum is a lower bound for S(x). In particular, calculation shows that, for x=199/200, the sum with k going from 0 to 5 is 0.50088158499... > 0.5008, so
S(199/200) > 0.5008.
Next note that S(x) = x - x^2 + S(x^4) > S(x^4). So
S((199/200)^(1/4)) > S(199/200) > 0.5008,
S((199/200)^(1/16)) > S((199/200)^(1/4)) > 0.5008,
S((199/200)^(1/64)) > S((199/200)^(1/16)) > 0.5008,
etc. Since the arguments of S approach 1, the limit can't be 1/2, so it doesn't exist.