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We know that the closed form of the generalized Euler sums involving harmonic number $$\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^k}}}} = \frac{1}{2}\left\{ {\left( {k + 2} \right)\zeta \left( {k + 1} \right) - \sum\limits_{i = 1}^{k - 2} {\zeta \left( {k - i} \right)\zeta \left( {i + 1} \right)} } \right\},$$ where $k\ge2$ is a positive integer and the harmonic number $H_n^{(k)}$ is defined by $$H^{(k)}_n:=\sum\limits_{j=1}^n\frac {1}{j^k}\quad {\rm and}\quad H^{(k)}_0:=0.$$ Next, we defined the generalized shifted harmonic number by $$H_\alpha ^{\left( m \right)} = \zeta \left( m \right) - \zeta \left( {m,\alpha + 1} \right),\;\alpha \notin N^-,\ 2\leq m\in N,$$ and for $m=1$, $${H_\alpha } \equiv H_\alpha ^{\left( 1 \right)} = \sum\limits_{k = 1}^\infty {\left( {\frac{1}{k} - \frac{1}{{k + \alpha }}} \right)} .$$ My question is that how to calculate the closed form of the following shifted harmonic number sums $$\sum\limits_{n = 1}^\infty {\frac{{{H_{n + \alpha }}}}{{{n^k}}}}=?$$ it can be expressed in terms of (Hurwitz) zeta values?

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  • $\begingroup$ For $~a=\tfrac12~$ we have $~H_{n+1/2}~=~2\cdot\displaystyle\sum_{k=0}^n\frac1{2k+1}~-~2\ln2.~$ $\endgroup$ – Lucian Apr 8 '17 at 18:26
  • $\begingroup$ In other words, $H_{n+1/2}~=~2~H_{2n+2}-H_{n+1}-2\ln2.$ $\endgroup$ – Lucian Apr 8 '17 at 18:34
  • $\begingroup$ A related question. $\endgroup$ – Lucian Apr 8 '17 at 20:09

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