The question we are considering concerns a sum over all permutation $\sigma \in S\_n$ (symmetric group) of a certain rational function:


$$\sum\_{\sigma \in S\_n} \frac{\sigma\_{j1} \sigma\_{j2}... \sigma\_{jk}}{\prod\_{i=1}^{n-1} (\sigma\_i - \sigma\_{i+1})} \stackrel{?}{=} 0,$$ with $k < n-1$ and $jm \in \{1,2,...,n\} \ \forall m \in \{1,..,k\}$. 


It is easy to see that the sum is equal to zero if there is a constant in the numerator by adding a term in the denominator, such that it becomes invariant under cyclic permutations, i.e.$$\sum\_{\sigma \in S\_n} \frac{1}{\prod\_{i=1}^{n-1} (\sigma\_i - \sigma\_{i+1})} = \sum\_{\sigma \in S\_n} \frac{(\sigma\_n - \sigma\_{1})}{\prod\_{i=1}^{n-1} (\sigma\_i - \sigma\_{i+1}) (\sigma\_n - \sigma\_{1})} = 0 \, .$$ The question is whether this sum is still zero if we include the product of at most $n-2$ elements of the form $\sigma\_{jm}$. The sum is unequal to zero for $k > n-2$.



sum over permutations equals zero?
Perhaps you wish to look at the paper



> 
> Tewodros Amdeberhan, *Explicit computations with the Divided Symmetrization operator*, Proc. Amer. Math. Soc., **144** no 7 (2016) pp 2799-2810, doi:[10.1090/proc/12931](https://doi.org/10.1090/proc/12931)
>  arXiv:[1406.0447](https://arxiv.org/abs/1406.0447)
> 
> 
> 


In particular, Proposition 0 is what you seek for.


