# The sum of difference products related to determinants with factorials

Let $$n \in \mathbb{N}$$ and $$a_0, a_1, \dots, a_n$$ be positive integers such that $$a_i\neq a_j$$ for $$i \neq j$$. Prove that

$$\sum_{0\leq k \leq n} \prod_{\substack{0\leq i \leq n\\i\neq k}} \frac{1}{a_i-a_k} = 0$$

The original problem was to evaluate the determinant of the matrix below: $$A =\, \begin{bmatrix} a_0!&(a_0+1)!&\cdots&(a_0+n)!\\ a_1!&(a_1+1)!&&\vdots\\ \vdots&&\ddots&\vdots\\ a_n!&\cdots&\cdots&(a_n+n)! \end{bmatrix}$$ and the determinant should be equal to $$\prod_{0\leq i \leq n}a_i! \prod_{0\leq i To prove this, I tried cofactor expansion and mathematical induction respect to $$n$$; \begin{align*} \mathrm{det}\,A &= \sum_{0\leq k \leq n} (-1)^k a_k! \prod_{\substack{0\leq i \leq n\\i\neq k}}(a_i+1)!\prod_{\substack{0\leq i and now it's done if the sum of the most right hand side is equal to $$1$$. I applied induction again to prove this; Let $$b_i=a_i+1$$ then the sum is \begin{align*} \sum_{0\leq k \leq n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{b_i}{b_i-b_k} &= \sum_{0\leq k < n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= \sum_{0\leq k < n}\frac{b_n}{b_n-b_k}\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= \sum_{0\leq k < n}\left(1+\frac{b_k}{b_n-b_k}\right)\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n} \end{align*} from the induction hypothesis, \begin{align*} &= 1 + \sum_{0\leq k < n}\frac{b_k}{b_n-b_k}\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= 1 + \prod_{0\leq i It'll be finally proven if the sum of most right hand side is $$0$$, and this is identical to the left hand side of the very first problem, except for $$a_i$$ is now $$b_i$$.

Later, I managed to prove the original problem in the different way, so the first problem is also proven. However I'm looking for more 'direct' proof not through a such tricky route. Does anyone have idea for this? Thank you.

• Are you acquainted with Cauchy matrices? Jul 29, 2020 at 17:12
• Indeed closely related (see the connection between my answer below and this one). Aug 5, 2020 at 6:00

In fact the same holds with the $$a$$'s being (distinct) elements of an abritrary field.
To prove it, we may assume $$n>0$$. Consider the partial fraction expansion $$\prod_{j=0}^n\frac{1}{x-a_j}=\sum_{i=0}^n\frac{A_i}{x-a_i},\quad A_i=\prod_{j\neq i}\frac{1}{a_i-a_j}$$ ($$A_i$$ are found by multiplying the identity by $$x-a_i$$, and letting $$x=a_i$$ then).
To get $$\sum_i A_i$$ out of it, an idea would be to multiply the identity by $$x$$ and take $$x\to\infty$$. But, in an abstract field, we can't do that. Instead, we put $$x=1/z$$ and divide the result by $$z$$: $$z^n\prod_{i=0}^n\frac{1}{1-a_i z}=\sum_{i=0}^n\frac{1}{1-a_i z}\prod_{j\neq i}\frac{1}{a_i-a_j}.$$ And it remains to put $$z=0$$.