Let ck:=∑j−0k(−1)j(kj)(1/ζ(2j+2)). We prove that the Riemann hypothesis is equivalent to ck≪k−3/4+ε for all ε>0; furthermore, we prove that ck≪k−3/4 implies that the zeros of ζ(s) are simple. This is closely related to M. Riesz's criterion which states that
the Riemann hypothesis is equivalent to ∑k=1∞((−1)k+1xk/(k−1)!ζ(2k))≪x1/4+ε as x→+∞, for all ε>0.