19 May
2005
19 May
'05
12:43 p.m.
Mike Speciner writes << 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
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.)] --Dan