From: Raymond Manzoni Subject: Re: Sum involving zeta and zeta' Date: Mon, 06 Dec 1999 19:41:36 +0100 Newsgroups: sci.math Keywords: Adamchik's formula for the Hurwitz zeta function Leroy Quet wrote: > > Let H(m)= 1+1/2+1/3+...1/m. > Let zeta(n)= sum_{k=1}^infinity [1/k^n]. > Let zeta'(n)= -sum_{k=1}^infinity [ln(k)/k^n]. (derivative of zeta) > Does: > sum_{m=1}^infinity [m (zeta'(m+1) +(zeta(m+1)-1) H(m))]= > zeta(2) +zeta'(2) ? > Thanks, > Leroy Quet Yes, There's probably a much nicer derivation but there are some rather interesting formulas here. We may use a formula concerning Hurwitz zeta for that purpose (V. Adamchik's ?) (*) sum{k=0}^infinity C(s+k-1,k)*zeta(s+k,a)*z^k = zeta(s,a-z) where Re(a)>0 and |z|<|a| If C(s+k-1,k) means gamma(s+k)/(gamma(s)*k!) then: sum{k=0}^infinity gamma(s+k)/(gamma(s)*k!)*zeta(s+k,a)*z^k = zeta(s,a-z) taking the derivative over s of gamma(s+k)/(gamma(s)*k!)*zeta(s+k,a) we get : [(gamma'(s+k)*gamma(s)-gamma(s+k)*gamma'(s))*zeta(s+k,a)+gamma(s+k)*gamma(s)*zeta'(s+k,a)] /(gamma(s)^2*k!) so that, back in the initial formula, and using gamma'(s)=gamma(s)*psi(s), we get: sum{k=0}^infinity C(s+k-1,k)*[psi(s+k)-psi(s))*zeta(s+k,a)+zeta'(s+k,a)]*z^k = zeta'(s,a-z) Let apply this for a=2, z=1 and s=2 then : sum{k=0}^infinity (k+1)*[(psi(2+k)-psi(2))*zeta(2+k,2)+zeta'(2+k,2)] = zeta'(2,1) since psi(2+k)-psi(2)=H_{k+1)-1, zeta(2+k,2)=zeta(2+k)-1, zeta'(2+k,2)=zeta'(2+k) and zeta'(2,1)=zeta'(2) we get : sum{k=0}^infinity (k+1)*[(H_{k+1)-1)*(zeta(2+k)-1)+zeta'(2+k)] = zeta'(2) setting k+1=m we get : sum{m=1}^infinity m*[(H_m-1)*(zeta(m+1)-1)+zeta'(m+1)] = zeta'(2) now since : sum{k=2}^infinity (zeta(k)-1)=1 and sum{k=2}^infinity k*(zeta(k)-1)=1+zeta(2) (and more generally, since you seem to like Stirling numbers, (**) sum{k=2}^infinity k^n*(zeta(k)-1) = 1+sum_{k=1}^n k!*S(n+1,k+1) where S(n,k) are Stirling number of the second kind. (*) and (**) were given in a paper of 1991 by Victor Adamchik (I think)) the second minus the first (and m=k-1) give : sum{m=1}^infinity m*(zeta(m+1)-1)=zeta(2) so that, adding both equations, we get: sum{m=1}^infinity m*[H_m*(zeta(m+1)-1)+zeta'(m+1)] = zeta(2)+zeta'(2) Hope that some of these formulas will interest you too, Raymond Manzoni