TITLE: Sum over all permutations
QUESTION [8 upvotes]: Playing with another question, I got this equality by probabilistic considerations. I guess there is a simple proof, but I'm not strong at this...
Let  $x_i >0$, $i=1,2 \cdots N$ 
Then
$$ \frac{1}{\displaystyle \prod_{i=1}^N  x_i}=\sum_{\sigma}\prod_{s=1}^N \frac{\displaystyle  1}{\displaystyle  \sum_{i=1}^s x_{\sigma(i)}} $$
where the sum is over all permutations of $\{ 1,2 \cdots N\}$
(I suspect that the assumption $x_i > 0$ is not necessary)

REPLY [5 votes]: Group the permutations into pairs with the same $\{\sigma(1),\sigma(2)\}$, and add their contributions.  From the first two factors, you get $$\frac{x_{\sigma(1)}+x_{\sigma(2)}}{x_{\sigma(1)}x_{\sigma(2)}(x_{\sigma(1)}+x_{\sigma(2)})}$$
and all the other factors are the same for that pair.  So you can replace $\frac1{x_{\sigma(1)}}\frac1{x_{\sigma(1)}+x_{\sigma(2)}}$ in every permutation by $\frac1{2x_{\sigma(1)}x_{\sigma(2)}}$, and you won't change the overall sum.
Now group the permutations into sets of six with the same $\{\sigma(1),\sigma(2),\sigma(3)\}$.  When you add their contributions, the first three factors simplify in the same way, so you get the same overall sum by replacing the first three factors by $\frac1{6x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)}}$.
And so on.