How would you calculate this sum? Let $x_1,...,x_n$ be $n$ distinct real numbers.
I'd like to prove that $$\sum_{j=1}^{n}
{
\frac
{1}
{\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}}
} = 0.$$
This makes me think of Cauchy determinants but I don't see any direct link. I tried to reduce to the same denominator but it led to nothing good. Any suggestion?
NB : Note that this sum equals $\sum_{k=1}^{n}{\frac{1}{Q'(x_k)}}$ where $Q = \prod_{k=1}^{n}{(X-x_k)}$.
Thank you!
 A: More generally, the identity holds also when $x_1,...,x_n$ are $n$ distinct complex numbers.
Consider the polynomial of degree at most $n-1$, given by
$$P(z):=\sum_{j=1}^{n}
{
\frac
{{\prod_{i=1 \land i \neq j}^{n}{(z-x_i)}}}
{\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}}
}-1.$$
It is easy to verify that $P(x_i)=0$ for $i=1,\dots,n$. Therefore, since $x_1,...,x_n$ are $n$ distinct numbers, the polynomial $P$ is identically zero. It follows that the coefficient of $z^{n-1}$ of $P$,  which is just
$$\sum_{j=1}^{n}\frac
{1}
{\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}},
$$
has to be zero.
A: Using the hint, we may rewrite
$$\sum_{k=1}^n \frac{1}{Q' (x_k)} = \lim_{h \to 0} \sum_{k=1}^n \frac{h}{Q(x_k + h)}$$
Where the above uses that $Q(x_k) = 0$. Then, getting a common denominator on the inside,
$$\lim_{h \to 0} \sum_{k=1}^n \frac{h}{Q(x_k + h)} = \lim_{h \to 0} \frac{h}{\prod_{i=1}^n Q (x_i + h)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h)$$
Now, note that the outside limit is simply
$$\frac{1}{\prod_{i=1}^n Q'(x_i)}$$
And $\prod_{i=1}^n Q'(x_i)$ is precisely the discriminant of $Q$ which is nonzero since $Q$ has distinct roots (up to a factor of $(-1)^{n(n-1)/2}$, but this doesn't change anything). Thus, we see:
$$\lim_{h \to 0} \frac{h}{\prod_{i=1}^n Q (x_i + h)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h) =\frac{1}{\prod_{i=1}^n Q'(x_i)} \lim_{h \to 0} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h)$$ $$ =   \frac{1}{\prod_{i=1}^n Q'(x_i)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i) = 0$$
Where we've used that each $x_i$ is a root of $Q$ in the above.
