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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asked Mar 22, 2019 at 18:42
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3$\begingroup$ @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, 2019 at 19:24
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$\begingroup$ @AlexM. but how does editing give reputation points? I thought that they are given either for questions or for answers. $\endgroup$– Fedor PetrovMar 22, 2019 at 20:12
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1$\begingroup$ @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, 2019 at 20:22
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$\begingroup$ What do you need from this sum? $\endgroup$– Alexey UstinovMar 29, 2019 at 11:20
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$\begingroup$ @AlexeyUstinov: I would like the RHS free of summation, but it is okay if it involves $H_n$. $\endgroup$– T. AmdeberhanMar 29, 2019 at 13:09
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2 Answers
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Denote the sum in question by $f(b,n)$, then $$\sum_{b,n\geq 0} f(b,n) y^b z^n = \frac{\log(1+y)\log(1-\frac{yz}{1-z})}{1-z-yz}.$$
answered Mar 22, 2019 at 21:00
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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 $$-2 \sum_{i=1}^n \frac{(-1)^i}{i^2} -\big( H_n\big)^2 -H_n^{(2)}.$$
