Timeline for A combinatorial identity involving harmonic numbers
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2017 at 21:22 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
deleted 4 characters in body
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| May 18, 2016 at 13:30 | comment | added | Max Alekseyev | This has reminded of my own proof of a similar identity: artofproblemsolving.com/community/q1h352146p1915602 | |
| May 18, 2016 at 8:35 | vote | accept | Chitsai Liu | ||
| Jun 3, 2017 at 1:29 | |||||
| May 17, 2016 at 16:32 | comment | added | esg | The following variant of the proof may be simpler: the relation $F(y)=[x^n] \frac{1}{1+x}\frac{1}{(1-\frac{x}{1+x}(1-y))^{n+1}}$ shows that $F(y)=(-1)^n F(1-y)$. Thus $F(1)-F(y)=(-1)^n \sum_{k=1}^n { n\choose k} { n+k \choose k} (-1)^{k+1} (1-y)^{k}$ and (using the Beta-integral $\int_0^1 (1-y)^{k-1} y^n\,dy=\frac{(k-1)!\,n!}{(n+k)!}$) $$\int_0^1 \frac{1 -F(y)}{1-y}\,y^n\,dy=(-1)^n \sum_{k=1}^n{n \choose k}\frac{(-1)^{k+1}}{k}=(-1)^n\,H_n$$ | |
| May 16, 2016 at 15:46 | history | answered | Fedor Petrov | CC BY-SA 3.0 |