Euler Sum Generalized = Sum With Stirling Number

Let S(r,y) be this generalization of an Euler sum,

S(r,y) =

---
\ H(k)
/ ---------,
--- (k+y)^r
k>=1

where H(k) is the kth harmonic number,
H(k) = sum{j=1 to k} 1/j.

(In linear-mode, S(r,y) = sum{k>=1} H(k)/(k+y)^r.)

As have said in earlier sci.math posts,

S(r,y) =

zeta(r+1,y)*r/2
+ H(y-1)*zeta(r,y)
-(1/2) sum{k=2 to r-1} zeta(k,y)*zeta(r+1-k,y),

where zeta(r,y) = sum{k>=0} 1/(k+y)^r.
(See:
http://mathworld.wolfram.com/HurwitzZetaFunction.html )



But now I post to share that S(r,y) =

---
\ S(k,r-1) *(H(k+y-1) - H(k-1))
(y-1)!/ -------------------------------,
--- k * (k+y-1)!
k>=1

where S(k,r-1) is an unsigned Stirling number of the first kind.
(ie sum{k>=0} S(k,m) x^k/ k! = (-ln(1-x))^m /m!.)

In linear-mode:

S(r,y)
= (y-1)! *sum{k>=1} S(k,r-1) *(H(k+y-1) - H(k-1))/(k*(k+y-1)!)

I get this as a direct consequence of:
S(r,y) =
(1/(r-1)!) *
integral{0 to 1} -ln(1-x) x^(y-1) (-ln(x))^(r-1) /(1-x) dx


Am I correct regarding all of the above?
I was not being rigorous when I found this.
And I am also error-prone lately.

thanks,
Leroy Quet

.

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