The following problem is likely too special for MO. However I have no clue how to deal with it, so I'll just try. Nevertheless it is a combinatorial problem and a discussion about general methods in this context will be interesting and helpful.

How can one determine whether the following equation

$\sum_{p=2}^n\frac{(-1)^p}{p!}\sum_{j_1+\cdots+j_p=n}^{j_k\geq 1} \frac{1}{j_1\cdot \cdots \cdot j_p}\sum_{1\leq l < m}j_m\cdot j_l =\frac{1}{2}$.

holds for any $n\in\mathbb{N}$ with $n\geq 2$? I checked it directly for $n\leq 6$, but couldn't find any pattern to start from.

Expressions like this arise frequently in the context of Lie infinity algebra morphisms and insight from professionals in combinatorics would be great.

P.S.: If someone has a better title feel free to change it.


One technique which is very helpful in proving identities like this one is to express that expression as a coefficient in a generating function.

Consider the expression $$\sum_{p\geq 2} \frac{(-1)^p}{p!}\binom{p}{2}\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right)^{p-2}\left(x^2+2x^3+3x^4+\cdots\right)$$ You can check that the coefficient of $x^n$ is exactly your expression. However one can use $$x^2+2x^3+3x^4+\cdots=\frac{x^2}{(1-x)^2}$$ and $$\frac{e^{-t}}{2}=\sum_{p\geq 2}\frac{(-1)^p}{p!}\binom{p}{2} t^{p-2}$$ to simplify this generating function to $$\frac{e^{\log(1-x)}}{2}\frac{x^2}{(1-x)^2}=\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{2}+\frac{x^5}{2}+\cdots,$$ and conclude that your expression is always equal to $\frac{1}{2}$.

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    $\begingroup$ Thanks Gjergji. I know the method of generating functions in principle. Since your answer came pretty fast, are you just very good at this or is there a systematic way to find the appropriate function? $\endgroup$ – Mark.Neuhaus Nov 9 '13 at 7:40

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