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In the middle of a physics calculation I have encountered a sum of the form:

$$\sum_{j=1}^n\bigg(\sum_{l=1\\ l\neq j}^n\dfrac{1}{x_j-x_l}\bigg)^2$$

The way the book proceeds implies that the sum of the terms which involve a product of two different fractions must vanish, that is, the following equality should hold:

$$\sum_{j=1}^n\sum_{l=1\\ l\neq j}^n\sum_{k=1\\ k\neq l,j}^n\bigg(\dfrac{1}{x_j-x_l}\cdot\dfrac{1}{x_j-x_k}\bigg)=0$$

Is there a way to prove this for arbitrary $n$? I tried doing it manually for $n=3$ and it vanishes indeed, but it is not trivial. I'm curious, any advice?

Wild Feather
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    Once you've proved it for the $n=3$ case, the general result follows. The sum goes over all triplets ${j,k,l}$ of distinct indices, and each triplet vanishes just like the $n=3$ case. – Jaap Scherphuis May 08 '23 at 12:27
  • @JaapScherphuis Oh, I think I see what you mean! But how can I write it formally? I would have:

    $$\sum_{j,k,l=1\ j\neq k\neq l}^n\dfrac{1}{x_j-x_k}\dfrac{1}{x_j-x_l}$$

    and this sum could be decomposed as the sum over permutations of triplets {j,k,l} where all indices are different?

     – Wild Feather May 08 '23 at 12:52
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    Yes, you could write it like that. Every unordered triplet $j\neq k\neq l$ is some permutation of an ordered triplet $j<k<l$. So your sum is equal to $$\sum_{j,k,l=1\ j<k<l}^n 2\left( \dfrac{1}{x_j-x_k}\dfrac{1}{x_j-x_l} + \dfrac{1}{x_k-x_l}\dfrac{1}{x_k-x_j} + \dfrac{1}{x_l-x_j}\dfrac{1}{x_l-x_k} \right)$$ The six terms are all possible permutations of the three chosen indices. Inside the sum we have the n=3 case which simplifies to zero. – Jaap Scherphuis May 08 '23 at 15:30

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