I have seen this formula, but how to prove this? $$2\sum\limits_{k=1}^\infty \frac{H_k}{\left( k+1 \right)^m} =m\zeta \left( m+1 \right)-\sum\limits_{k=1}^{m-2}{\zeta \left( m-k \right)\zeta \left( k+1 \right)}$$
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$\zeta(m) = \sum_{k \ge 1} k^{-m}$, $H_n = \sum_{1 \le k \le n} k^{-1}$, so I'd try something like summation by parts. – vonbrand Feb 03 '13 at 02:30
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1http://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk/275657#275657 deals with something very similar. – Feb 03 '13 at 02:49