From: Raymond Manzoni <raymman@club-internet.fr>
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