For each positive integer k, let ak(n)=(∑pp−s)k=∞∑n=1ak(n)n−s,
where σ=Re(s)>1, and the sum on the left runs over all primes p. This paper is devoted to proving the following theorem: If 1/2<σ<1, then maxk(∑n≤Nak(n)2n−2σ)1/2k≈(logN)1−σ/loglogN
and (∞∑n=1ak(n)2n−2σ)1/2k≈k1−σ/(logk)σ.
The constants implied by the ≈ sign may depend upon σ. This theorem has applications to the Riemann zeta function.