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] As far as characterizing these series, it might seem weird but true that the only thing that matters is the series' behavior "at infinity". 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). 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. --ms Gareth McCaughan wrote:
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.
Neat!
This puts the kibosh to what I thought was a proof that if the alternating series sum k=0 to oo of (-1)^k x^(n_k) converges, then it converges to 1/2 (as Gareth asked).
[The "proof" just looked at the terms of the Cesaro summation, which for a convergent series also converges. That is, the averages of the first n terms of the given series for n = 1,2,3,.... The sequence of these averages' limits as x -> oo approaches 1/2. (But I see from Mike counterexample that one can't argue this implies the conclusion I wanted.)]
I was wondering about that approach too, but being out of practice and busy with other things I hadn't had time to try to convince myself either that it works or that it doesn't. :-)
So, call a sequence "very good" if the limit exists *and equals 1/2*. It seems that lots of sequences are very good. Is there any hope of characterizing them? (I rather suspect not.)