On Thursday 19 May 2005 21:04, Mike Speciner wrote:
Of course, 0 j k k+j 2k 2k+j ... (with j < k) gives (1-x^j)/(1-x^k) which is (1+x+x^2+ ... +x^(j-1))/(1+x+x^2+ ... +x^(k-1)), giving a limit of j/k, so you can get any rational limit in the open interval (0,1). Clearly, you can't have a limit < 0, because pairs of successive terms always have positive sum. Similarly the limit can't be > 1 (don't pair the first term, but pair the rest). But how about =0? =1? What about irrational limits? [Does 2 3 4 5 8 9 16 17 32 33... give 0 and thus 0 2 3 4 5 8 9 16 17 32 33 ... give 1? I think the former is < 1/2^k for all k]
Suppose we have a sequence of the form n1 ni+j n2 n2+j n3 n3+j ... where successive n's differ by more than j. Then f(x) = (1-x^j) (x^n1 + x^n2 + x^n3 + ...) = (1-x^j) g(x), say f(exp(-h)) = (jh+O(h^2)) g(exp(-h)) which will tend to jt (say) if g(exp(-h)) = t/h + o(1/h). So, how fast does g(x) grow as x->1? I bet this is very well understood by those who understand such things; but we can get a very crude estimate as follows. By truncating after a term x^k, we obviously don't increase the sum, and we decrease it by at most x^(k+1)/(1-x). So, suppose that the first M(h) terms are at least u and the terms after the N(h)th are at most v. Then the sum is at least M(h)u and at most N(h)u+v/h + o(1/h). Consider, for instance, your example where the exponents in g are the powers of 2; take u=1/e and v=1/e^n for some n. Then the k'th term is exp(-2^k.h). That's >=u as long as k <= log_2(1/h), and <=v as long as k >= n log_2(1/h). We're only interested in the upper bound, which says that g(exp(-h)) <= n/e log_2(1/h) + 1/(h.e^n) + o(1/h), so that f(exp(-h)) <= n/e h log_2(1/h) + e^-n + o(1). Since h log h -> 0 as h -> 0, this provides an arbitrarily small (positive) upper bound on "f(1)", so your conjecture is right. I suspect a lot of fat could be trimmed from this proof. Hmm. For an arbitrary sequence, we can write f(x) = (1-x) g(x) where g(x) is a sum of distinct powers of x. Now the following might be a theorem: Let A be a subset of the integers, and suppose it has density r in the sense that #{a in A : a<n}/n -> r as n -> oo. Then sum { a in A } of (1-h)^a = r/h + o(1/h) as h->0. If so, then it's easy to construct sequences with any limit in [0,1]. I suspect it's not only a theorem but an easyish one, but it's late (local time) and this message is long enough already.
As far as characterizing these series, it might seem weird but true that the only thing that matters is the series' behavior "at infinity".
I'd have said "obvious" rather than "weird but true" :-).
Altering any finite part of the series has no effect on the limit. Also, adding a constant to all the exponents in a series (i.e. multiplying by x^C) has no effect on the limit, nor does multiplying all the exponents by a constant (which just substitutes x^C for x).
Right, except that ... If by "altering any finite part" you mean "splicing out some finite subsequence and replacing it with another", then your first statement is *almost* true -- inserting or removing one term applies t -> 1-t to the limit.
Having said that, it seems like a way to get an arbitrary real between 0 and 1 might be to exhibit an increasing sequence of rationals (with each denominator a multiple of the previous denominator) converging to that real and produce the sequence which includes pairs of terms for each of those rationals j/k (as described at the beginning of this email, but starting with the k k+j pair) until the k' k'+j' pair from the next rational.
Sounds promising. Got a proof? :-) -- g