**bts@unc.UUCP** (10/19/83)

Here's a question concerning enumerations of the rational numbers and divergent series. I don't know the answer, and I've never seen this question asked before. (If you've seen it elsewhere, please provide references as best you can.) And, I don't have any applications in mind, so try solving it for fun only. It's possible to enumerate the set of rational numbers (see any textbook on analysis or set theory) as a sequence q1, q2, q3,... , with or without repetitions. Also, given any sequence of numbers a1, a2, a3,... it's possible to find a sequence b1, b2, b3,... so that the a's are the sequence of partial sums of the infinite series of b's, b1+b2+b3+b4+..., i.e. a1 = b1 a2 = b1 + b2 ... an = b1 + b2 + ... + bn ... Just let b1 = a1, and after that bn = an-a(n-1). If the a's are rational numbers, clearly the b's are rational, also. Here's the question: Is it possible to have two enumerations of the rational numbers q1, q2, q3 ,... and r1, r2, r3 ,... so that the r's are the sequence of partial sums of the infinite series of q's, q1+q2+q3+... ? Is it possible to do this so that there are no repetitions in either sequence? If so, can you give the enumeration effec- tively, or does your proof rely on non-constructive methods (e.g. the Axiom of Choice)? Bruce Smith, UNC-Chapel Hill decvax!duke!unc!bts (USENET) bts.unc@udel-relay (other NETworks)

**bts@unc.UUCP (Bruce Smith)** (10/19/83)

In my query on enumerating the rational numbers (unc.6029, "A question on enumerating the rationals"), feel free to have the sequences enumerate the non-zero rationals, instead of the whole set. Having 0 as a term in a series means there'll be at least one repeat in its sequence of partial sums. Bruce Smith, UNC-Chapel Hill

**leichter@yale-com.UUCP (Jerry Leichter)** (10/20/83)

There is a trivial answer of "no" to the existence of an enumeration of the rationals, without repetition, whose partial sums are also an EWOR: 0 must occur in the enumeration, forcing a repetition in the partial sum at that point. As to the general question: Looks much harder. I'd guess "yes", based on gut feel only. I'd like to see a proof one way or the other, though. -- Jerry decvax!yale-comix!leichter leichter@yale