Volume 5 - 1982


1. On a question of Ramachandra

Hugh L Montgomery.
For each positive integer k, let ak(n)=(pps)k=n=1ak(n)ns,
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(nNak(n)2n2σ)1/2k(logN)1σ/loglogN
and (n=1ak(n)2n2σ)1/2kk1σ/(logk)σ.
The constants implied by the sign may depend upon σ. This theorem has applications to the Riemann zeta function.