I just had to make use of an elementary rational function identity (below). The proof is a straightforward exercise, but that isn't the point. First, "my" identity is almost surely
not original, but I don't have a reference for it. Perhaps someone knows it (like a lost cat without a collar) or, more likely, could spot this as a special case of a more general identity. Second, the obvious proof is not much of an explanation: a combinatorial identity often arises for a conceptual reason, and I'd be happy to hear if anyone sees mathematics behind this one. 


Let $f(x\_1,\ldots,x\_n)=\prod\_{p=1}^n\big(\sum\_{i=p}^n x\_i\big)^{-1}$. Then
$$
f(x\_1,\ldots,x\_n)+f(x\_2,x\_1,x\_3,\ldots,x\_n)+\cdots+f(x\_2,\ldots,x\_n,x\_1)=\big(\sum\_{i=1}^n x\_i\big)/x\_1\cdot f(x\_1,\ldots,x\_n),
$$
where $x\_1$ appears as the $i$th argument to $f$ in the $i$th summand on the left side, for $1\leq i\leq n$. But why?



rational function identity
I have seen a cat of a similar breed in the representation theory of symmetric groups. Out of habit, let me quote a lemma attributed to Littlewood in


Donald Knutson, *$\lambda$-rings and the Representation Theory of the Symmetric Group*, Springer 1973 (LNM #308), Chapter III, section 2, p. 149:


$\sum\limits\_{\sigma\in S\_n} f\left(x\_{\sigma\left(1\right)},x\_{\sigma\left(2\right)},...,x\_{\sigma\left(n\right)}\right) = \frac{1}{x\_1x\_2...x\_n}$.


At the moment, neither does this cat imply yours, nor the other way round. But can we cross them?


Let me try. The left paw side of your cat is $\sum\limits\_{\sigma\in \mathrm{Sh}\left(1,n-1\right)} f\left(x\_{\sigma^{-1}\left(1\right)},x\_{\sigma^{-1}\left(2\right)},...,x\_{\sigma^{-1}\left(n\right)}\right)$, where $\mathrm{Sh}\left(a,b\right)$ is defined as the subgroup


$\left\lbrace \sigma \in S\_{a+b} \mid \sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(a\right) \text{ and } \sigma\left(a+1\right) < \sigma\left(a+2\right) < ... < \sigma\left(a+b\right) \right\rbrace$


of the symmetric group $S\_{a+b}$. (The elements of this subgroup $\mathrm{Sh}\left(a,b\right)$ are known as *$\left(a,b\right)$-shuffles*.) Now I suspect tat


$\sum\limits\_{\sigma\in \mathrm{Sh}\left(a,b\right)} f\left(x\_{\sigma^{-1}\left(1\right)},x\_{\sigma^{-1}\left(2\right)},...,x\_{\sigma^{-1}\left(a+b\right)}\right) = f\left(x\_1,x\_2,...,x\_a\right) f\left(x\_{a+1},x\_{a+2},...,x\_{a+b}\right)$


for any $a$ and $b$ and any $x\_i$.


This generalizes your cat. Does it generalize Littlewood's? Yes, at least if we generalize it even further, to the so-called *$\left(a\_1,a\_2,...,a\_k\right)$-multishuffles* (which are permutations $\sigma\in S\_{a\_1+a\_2+...+a\_k}$ increasing on each of the intervals $\left[a\_i+1,a\_{i+1}\right]$, where $a\_0=0$ and $a\_{k+1}=n$). This is not much of a generalization, since it follows from the $\left(a,b\right)$-shuffle version by induction over $k$, but applying it to $\left(1,1,...,1\right)$-multishuffles (which are simply all the elements of $S\_n$) yields Littlewood's cat.


Now I see that Littlewood's cat even follows from yours, if we notice that every permutation $\sigma\in S\_n$ can be written uniquely as a product $t\_1t\_2...t\_{n-1}$, where each of the $t\_k$ moves the $k$ some places to the right. (This is one of the stupid sorting algorithms.)


Oh, and I don't have a proof of my cat, but it can catch mice, so it's a good cat, isn't it?


