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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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  • $\begingroup$ $\zeta(m) = \sum_{k \ge 1} k^{-m}$, $H_n = \sum_{1 \le k \le n} k^{-1}$, so I'd try something like summation by parts. $\endgroup$ – vonbrand Feb 3 '13 at 2:30
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    $\begingroup$ math.stackexchange.com/questions/275643/… deals with something very similar. $\endgroup$ – user17762 Feb 3 '13 at 2:49