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Let $H_k=\sum_{j=1}^k\frac1j$ be the harmonic numbers.

QUESTION. Can you find an evaluation of the following sum? $$\sum_{a=1}^b(-1)^a\binom{n}{b-a}\frac{H_{b-a}}a.$$

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    $\begingroup$ @user64494, this is the last warning: you have currently gained 100 reputation points from 50 edits performed since the 6th of March, the vast majority of which change the capitalization of the title or perform similarly minor edits. If you don't stop right now, I'll flag your behaviour to a moderator, and open a thread on Meta about this. You are obviously abusing the system in order to gain reputation and badges. Please understand that MO is not a game! It is also because of people like you that valuable former members (including Fields medalists) stopped using MO. $\endgroup$ – Alex M. Mar 22 '19 at 19:24
  • $\begingroup$ @AlexM. but how does editing give reputation points? I thought that they are given either for questions or for answers. $\endgroup$ – Fedor Petrov Mar 22 '19 at 20:12
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    $\begingroup$ @FedorPetrov: In order to encourage users to get involved in the curating of the site, users with <2000 points get +2 points for each approved edit. Once they get past 2000, their edits no longer need approval, but they also stop receiving those +2. This is a SE-wide policy, not a MO-only one. I, for one, would gladly drop it, though - at least on the more "academic" communities, like MO. $\endgroup$ – Alex M. Mar 22 '19 at 20:22
  • $\begingroup$ What do you need from this sum? $\endgroup$ – Alexey Ustinov Mar 29 '19 at 11:20
  • $\begingroup$ @AlexeyUstinov: I would like the RHS free of summation, but it is okay if it involves $H_n$. $\endgroup$ – T. Amdeberhan Mar 29 '19 at 13:09
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Denote the sum in question by $f(b,n)$, then $$\sum_{b,n\geq 0} f(b,n) y^b z^n = \frac{\log(1+y)\log(1-\frac{yz}{1-z})}{1-z-yz}.$$

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Using the summation package Sigma one can discover (and prove) the right hand of $$ \sum_{a=1}^b \frac{(-1)^a \binom{n}{-a +b } H_{-a +b }}{a}=\binom{n}{b} \left[ \big( -H_n +H_{-b +n } \big) H_b - \sum_{i=1}^b \frac{(-1)^i}{i^2 \binom{n}{i}} \right].$$ Note that the arising hypergeometric sum on the right-hand side cannot be simplified further using, e.g., only harmonic numbers (or similar type of special functions). However, if one sets $b=n$, the right hand-side boils down to $$-2 \sum_{i=1}^n \frac{(-1)^i}{i^2} -\big( H_n\big)^2 -H_n^{(2)}.$$

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