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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  • $\begingroup$ What do you need from this sum? $\endgroup$ Commented Mar 29, 2019 at 11:20
  • $\begingroup$ @AlexeyUstinov: I would like the RHS free of summation, but it is okay if it involves $H_n$. $\endgroup$ Commented Mar 29, 2019 at 13:09

2 Answers 2


Denote the sum in question by $f(b,n)$, then $$\sum_{b,n\geq 0} f(b,n) y^b z^n = \frac{\log(1+y)\log(1-\frac{yz}{1-z})}{1-z-yz}.$$


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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