I found the following formula
∞∑n=1Hnnq=(1+q2)ζ(q+1)−12q−2∑k=1ζ(k+1)ζ(q−k)
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
k∑j=0ζ(k+2−j)ζ(j+2)=∞∑m=1∞∑n=1k∑j=01mk+2−jnj+2=(k+1)ζ(k+4)+∞∑m,n=1m≠n1m2n21mk+1−1nk+11m−1n=(k+1)ζ(k+4)+∞∑m,n=1m≠n1nmk+2(n−m)−1mnk+2(n−m)=(k+1)ζ(k+4)+2∞∑m=1∞∑n=m+11nmk+2(n−m)−1mnk+2(n−m)=(k+1)ζ(k+4)+2∞∑m=1∞∑n=11(n+m)mk+2n−1m(n+m)k+2n=(k+1)ζ(k+4)+2∞∑m=1∞∑n=11mk+3n−1(m+n)mk+3−2∞∑m=1∞∑n=11m(n+m)k+3+1n(n+m)k+3=(k+1)ζ(k+4)+2∞∑m=1Hmmk+3−4∞∑n=1∞∑m=11n(n+m)k+3=(k+1)ζ(k+4)+2∞∑m=1Hmmk+3−4∞∑n=1∞∑m=n+11nmk+3=(k+1)ζ(k+4)+2∞∑m=1Hmmk+3−4∞∑n=1∞∑m=n1nmk+3+4ζ(k+4)=(k+5)ζ(k+4)+2∞∑m=1Hmmk+3−4∞∑m=1m∑n=11nmk+3=(k+5)ζ(k+4)+2∞∑m=1Hmmk+3−4∞∑m=1Hmmk+3=(k+5)ζ(k+4)−2∞∑m=1Hmmk+3
Letting q=k+3 and reindexing j↦j−1 yields
q−2∑j=1ζ(q−j)ζ(j+1)=(q+2)ζ(q+1)−2∞∑m=1Hmmq
and finally
∞∑m=1Hmmq=q+22ζ(q+1)−12q−2∑j=1ζ(q−j)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(n−m)+1mnk+2(m−n)
0(5) n↦n+m and change the order of summation
0(6) 1mn=1m(m+n)+1n(m+n)
0(7) Hm=∑∞n=11n−1n+m and use the symmetry of 1m(n+m)k+3+1n(n+m)k+3
0(8) m↦m−n
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
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