Generalized Euler sum ∑∞n=1Hnnq\sum_{n=1}^\infty \frac{H_n}{n^q}

I found the following formula

n=1Hnnq=(1+q2)ζ(q+1)12q2k=1ζ(k+1)ζ(qk)

and it is cited that Euler proved the formula above , but how ?

Do there exist other proofs ?

Can we have a general formula for the alternating form

n=1(1)n+1Hnnq

Answer

kj=0ζ(k+2j)ζ(j+2)=m=1n=1kj=01mk+2jnj+2=(k+1)ζ(k+4)+m,n=1mn1m2n21mk+11nk+11m1n=(k+1)ζ(k+4)+m,n=1mn1nmk+2(nm)1mnk+2(nm)=(k+1)ζ(k+4)+2m=1n=m+11nmk+2(nm)1mnk+2(nm)=(k+1)ζ(k+4)+2m=1n=11(n+m)mk+2n1m(n+m)k+2n=(k+1)ζ(k+4)+2m=1n=11mk+3n1(m+n)mk+32m=1n=11m(n+m)k+3+1n(n+m)k+3=(k+1)ζ(k+4)+2m=1Hmmk+34n=1m=11n(n+m)k+3=(k+1)ζ(k+4)+2m=1Hmmk+34n=1m=n+11nmk+3=(k+1)ζ(k+4)+2m=1Hmmk+34n=1m=n1nmk+3+4ζ(k+4)=(k+5)ζ(k+4)+2m=1Hmmk+34m=1mn=11nmk+3=(k+5)ζ(k+4)+2m=1Hmmk+34m=1Hmmk+3=(k+5)ζ(k+4)2m=1Hmmk+3
Letting q=k+3 and reindexing jj1 yields
q2j=1ζ(qj)ζ(j+1)=(q+2)ζ(q+1)2m=1Hmmq
and finally
m=1Hmmq=q+22ζ(q+1)12q2j=1ζ(qj)ζ(j+1)


Explanation

0(1) expand ζ
0(2) pull out the terms for m=n and use the formula for finite geometric sums on the rest
0(3) simplify terms
0(4) utilize the symmetry of 1nmk+2(nm)+1mnk+2(mn)
0(5) nn+m and change the order of summation
0(6) 1mn=1m(m+n)+1n(m+n)
0(7) Hm=n=11n1n+m and use the symmetry of 1m(n+m)k+3+1n(n+m)k+3
0(8) mmn
0(9) subtract and add the terms for m=n
(10) combine ζ(k+4) and change the order of summation
(11) Hm=mn=11n
(12) combine sums

Attribution
Source : Link , Question Author : Zaid Alyafeai , Answer Author : robjohn

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