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.



Combinatorial Technique Needed
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 $\frac{1}{2}$.


