On chagnes of signs in infinite series

(1 .2) E(k)ak = Sn,j k=1 k = j (modn!) for n=1, 2,. .. and O~j-n!-1. When sn , j =0 for all n and j, (1 .1) follows automatically. Since every arithmetic progression with modulus m is a disjoint union of (m-1)! arithmetic progressions with modulus m!, Theorem 1 follows from Theorem 2. Also, by using this argument and (1 .2), we see that in Theorem 2 each series Z E (n) an is, n=b(mod m) in fact, prescribed for all b and m, 0-_b-_m-1. It is clear that conditions (1 .1) on {sn, j } are necessary. The main purpose of this paper is to prove the following results. Theorem 1. Let {an } be a sequence of positive real numbers monotonically decreasing to zero such that Z a n =-. Then there exist signs E(n)=±1, n = 1, 2,. . ., such that for every integer m-1 and every integer b with O-b-m-1 2' E(n)a n = 0. n=b (mod m) The above theorem will be a trivial consequence of the following result which is the main theorem in this paper. Theorem 2. Let {an } be a sequence as in Theorem 1 and let sn , j , n=1, 2,. .. and j=0,. . .,n!-1 be real numbers satisfying the conditions n!-1 (1 .1) j=0 j=d (mod (n-1)!)