In some work on QFT the following identity has come up:
$$
\sum\_{\sigma \in S\_n}\sum\_{j=1}^n \left(\sum\_{l=1}^j w\_{\sigma\_l}\right)\prod\_{i=1,i\neq j}^{n}\frac{1}{\sum\_{l=1}^j z\_{\sigma\_l}-\sum\_{l=1}^i z\_{\sigma\_l}}=0
$$
for $2<n$ where $w\_1,...,w\_n$ are arbitrary numbers and $z\_1,...,z\_n$ are such that
$$
\sum\_{z\in S}z\neq 0,\ \ \forall S\subsetneq \{z\_1,...,z\_n\}
$$
I have verified this symbolically for $2<n<6$ and using specific values for $w, z$ I have verified it up to $n=8$. Do you have any ideas on how to approach a proof?


An idea I've thought of so far: We could write a function 
$$
f(z)=\sum\_{\sigma\in S\_n}\prod\_{i=1}^{n}\frac{1}{z-\sum\_{l=1}^i z\_{\sigma\_l}}
$$
The integral of $f(z)$ around $\infty$ is then clearly 0 since it dies off fast enough. We can also express this integral as a sum over residues and we find almost the above identity but without the weights $w\_l$. In this case the identity is true also without the sum over permutations but somehow the sum over permutations allows for the weights $w\_l$.


For convenience, here is a mathematica expression of the identity:



```
Sum[
    Sum[
        Sum[w[p[[l]]], {l, 1, j}]
            Product[
                If[i == j,
                    1, 
                    1/(Sum[z[p[[l]]], {l, 1, j}] - Sum[z[p[[l]]], {l, 1, i}])],
                {i,n}],
            {j, n}],
        {p, Permutations[Range[n]]}]

```


Identity involving sum over permutations
**June 18, 2019 Edit: The identity is now proved and generalized. See below.**


I did not have time to look at your identity for very long but I am pretty sure it follows from Lemma II.2 in my article ["Trees, forests and jungles: a botanical garden for cluster expansions"](https://arxiv.org/abs/hep-th/9409094) with Vincent Rivasseau.


A starting point. Make a set of $n+1$ elements by adding a root numbered say $0$. Define the edge variables $u\_{ij}$ $(i<j)$ to be zero except $u\_{0,i}=z\_i$. An ordered forest of length $\tau$, i.e., $(\{i\_1,j\_1\},\{i\_2,j\_2\},\ldots,\{i\_{\tau},j\_{\tau}\})$ which contributes in Lemma II.2 namely for which 
$$
\prod\_{\nu=1}^{\tau}u\_{i\_{\nu},j\_{\nu}}\neq 0
$$
basically is sequence of distinct elements $(j\_1,\ldots,j\_{\tau})$ in $\{1,2,\ldots,n\}$. And for maximal length $\tau=n$, this is just a permutation $\sigma$.




---


Not quite there yet, although it looks promising.


I use formula (II.6) in Lemma II.2 of my article for the set $\{0,1,\ldots,n\}$ instead of $\{1,\ldots,n\}$. In the notations of the paper, I define the edge variables $u\_{\{ij\}}$ by $u\_{\{0i\}}=z\_i$ for $1\le i\le n$ and zero otherwise. I also define the edge variables $v\_{\{ij\}}$ by $v\_{\{0i\}}=y\_i$ for $1\le i\le n$ and $1$ otherwise. I will rename $\nu$ from the paper and now call it $j$. I will also rename $\mu$ from the paper and now call it $i$. An ordered forest of length $\tau$ is a sequence of edges $(l\_1,\ldots,l\_{\tau})$. In order to contribute it has to be of the form $l\_a=\{0,\sigma(a)\}$ for $\sigma=(\sigma(1),\ldots,\sigma(\tau))$ a sequence of distinct elements of $\{1,2,\ldots,n\}$ of length $\tau$.
With this harmonization of notations, Lemma II.2 from the paper gives
$$
y\_1\cdots y\_n=\sum\_{\tau=0}^{n}\sum\_{\sigma}z\_{\sigma(1)}\cdots z\_{\sigma(\tau)}
\sum\_{j=0}^{\tau}\frac{y\_{\sigma(1)}\cdots y\_{\sigma(j)}}{\prod\_{i=0,i\neq j}^{\tau}
(Z\_{j}^{\sigma}-Z\_{i}^{\sigma})}
$$
where for $k=0,1,\ldots,\tau$, $Z\_{k}^{\sigma}=z\_{\sigma(1)}+\cdots+z\_{\sigma(k)}$.
Might as well define also for $k=0,1,\ldots,\tau$, $W\_{k}^{\sigma}=w\_{\sigma(1)}+\cdots+w\_{\sigma(k)}$.


I will then carefully separate the cases where $i$ or $j$ are zero from the rest.


$$
y\_1\cdots y\_n=\sum\_{\tau=0}^{n}\sum\_{\sigma}z\_{\sigma(1)}\cdots z\_{\sigma(\tau)}
\sum\_{j=1}^{\tau}\frac{y\_{\sigma(1)}\cdots y\_{\sigma(j)}}{Z\_{j}^{\sigma}\prod\_{i=1,i\neq j}^{\tau}
(Z\_{j}^{\sigma}-Z\_{i}^{\sigma})}
$$
$$
+\sum\_{\tau=0}^{n}\sum\_{\sigma}z\_{\sigma(1)}\cdots z\_{\sigma(\tau)}
\frac{1}{\prod\_{i=1}^{\tau}
(-Z\_{i}^{\sigma})}
$$
Now I set $y\_i=e^{sz\_i+tw\_i}$. I take $\frac{\partial^2}{\partial s\partial t}$ and set $s=t=0$.
This gives the identity
$$
(w\_1+\cdots+w\_n)(z\_1+\cdots+z\_n)=
\sum\_{\tau=0}^{n}\sum\_{\sigma}z\_{\sigma(1)}\cdots z\_{\sigma(\tau)}
\sum\_{j=1}^{\tau}\frac{W\_{j}^{\sigma}}{\prod\_{i=1,i\neq j}^{\tau}
(Z\_{j}^{\sigma}-Z\_{i}^{\sigma})}
$$
Now I am stuck. I would like to "take the coefficient" of $z\_1\cdots z\_n$ which would somehow force $\tau=n$ and $\sigma$ to be a permutation but I don't see it yet. 




---


It's a bit late for me so can't give details but I think I know how to peal off $\tau=n$. Use Möbius inversion in the (Boolean) poset of subsets $I=\{i\_1<\cdots<i\_k\}$ of $[n]=\{1,\ldots,n\}$.
Namely apply the identity I just proved not just for the whole set $\{1,\ldots,n\}$ and corresponding variables, but also for all the $I$'s.


Really really have to go but basically the wanted result amounts to showing
$$
\sum\_{I\subset [n]} (-1)^{n-|I|}\left(\sum\_{i\in I} w\_i\right)
\left(\sum\_{i\in I} z\_i\right)=0
$$
when $n\ge 3$.




---


OK here is the endgame.


I will denote by $|\cdot|$ the cardinality of a finite set. For $I\subset[n]$, let
$$
B\_I=\sum\_{\sigma}z\_{\sigma(1)}\cdots z\_{\sigma(|I|)}
\sum\_{j=1}^{|I|}\frac{W\_{j}^{\sigma}}{\prod\_{i=1,i\neq j}^{|I|}
(Z\_{j}^{\sigma}-Z\_{i}^{\sigma})}
$$
where the sum is over sequences $\sigma=(\sigma(1),\ldots,\sigma(|I|))$ which form a permutation of the elements of $I$.
Also define
$$
A\_I=\sum\_{J\subset I}B\_J\ .
$$
Then the previous identity applied to the variables in $I$ instead of the whole set $[n]$, give
$$
A\_I=\left(\sum\_{a\in I} w\_a\right)
\left(\sum\_{b\in I} z\_b\right)\ .
$$
By Möbius inversion in the Boolean lattice, we get
$$
B\_I=\sum\_{J\subset I}(-1)^{|I|-|J|}A\_J\ .
$$
The LHS in the OP's identity is $B\_{[n]}$ divided by $z\_1\cdots z\_n$.
However,
$$
B\_{[n]}=\sum\_{J\subset[n]}(-1)^{n-|J|}
\left(\sum\_{a\in J} w\_a\right)
\left(\sum\_{b\in J} z\_b\right)
$$
$$
=\sum\_{a,b\in [n]}w\_a z\_b\sum\_{J,\{a,b\}\subset J\subset[n]}
(-1)^{n-|J|}=\sum\_{a,b\in [n]}w\_a z\_b(1-1)^{n-|\{a,b\}|}=0
$$
by Newton's binomial theorem and the hypothesis $n\ge 3$. QED


Note that the OP's identity admits a trivial generalization to families of weights $w^{(1)},\ldots,w^{(k)}$, namely
$$
\sum\_{\sigma\in\mathfrak{S}\_n}
\sum\_{j=1}^{n}\frac{W\_{j}^{(1),\sigma}\cdots W\_{j}^{(k),\sigma}}{\prod\_{i=1,i\neq j}^{n}
(Z\_{j}^{\sigma}-Z\_{i}^{\sigma})}\ =\ 0
$$
as soon as $n\ge k+2$. Here I used the generalized notation $W\_{j}^{(r),\sigma}=w\_{\sigma(1)}^{(r)}+\cdots+w\_{\sigma(j)}^{(r)}$.


Finally more information about identities like in Lemma II.2 of my article with Rivasseau can be found in my [dedicated webpage](http://people.virginia.edu/~aa4cr/forestpage.html) (in dire need of updates though). 


