Showing that $\sum_{i = 1}^m \frac{1}{\prod_{j = 1, j \neq i}^m (a_j - a_i)}$ is zero Question: How can I prove, for $m \geq 2$ and reals $a_1 < a_2 < \dots < a_m$ that $$\sum_{i = 1}^m \frac{1}{\prod_{j = 1, j \neq i}^m (a_j - a_i)} = 0?$$
Context: In Gamelin's text on Complex Analysis, exercise VII.6.4 asks to prove that $$\text{PV}\int_{-\infty}^\infty \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = 0,$$ which can be done using a contour integral around a half disk $\partial D$ (of radius $R$) in the upper half-plane, with small semicircular indents (of radius $\varepsilon$) above the singularities $a_1, a_2, \dots, a_m$ on the real axis. The method is straightforward, but when applying the fractional residue theorem to the semicircle indents $\gamma_b$, the sum of their contributions becomes $$\sum_{b = 1}^m \lim_{\varepsilon \rightarrow 0} \int_{\gamma_b} \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = \sum_{b = 1}^m \frac{-\pi i}{\prod_{j = 1, j \neq b}^m (a_j - a_b)},$$ and it is easy to show (using the ML-estimate) that the contribution of the integral over the semicircumference is negligible as $R \to \infty$. Thus, by Cauchy's Theorem, we have that $$\lim_{\varepsilon \to 0, R \to \infty}\int_{\partial D} \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = \sum_{b = 1}^m\frac{-\pi i}{\prod_{j = 1, j \neq b}^m (a_j - a_b)} + \text{PV} \int_{-\infty}^\infty \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = 0,$$ which gives the result the question wants, if the identity I am trying to prove is true.
I convinced myself the identity holds by trying for small values of $m$, but I have yet to come up with any rigorous proof. I have attempted an induction argument, but I am having trouble constructing the inductive step. Any hints/advice would be greatly appreciated.
 A: Let $\displaystyle\;P(\lambda) = (\lambda - a_1)(\lambda - a_2) \cdots (\lambda - a_m) = \prod_{i=1}^m (\lambda - a_i)$.
By product rule, we have
$$P'(\lambda) 
= {\small\begin{align}
     & (\lambda - a_1)'(\lambda - a_2)\cdots(\lambda - a_m)\\
   + & (\lambda - a_1)(\lambda - a_2)' \cdots (\lambda - a_m ) \\
   + & \cdots\\
   + & (\lambda - a_1)(\lambda - a_2) \cdots (\lambda - a_m)'
  \end{align}}
= \sum_{i=1}^m (\lambda - a_i)'\prod_{j=1,\ne i}^m (\lambda - a_j)
= \sum_{i=1}^m \prod_{j=1,\ne i}^m (\lambda - a_j)
$$
RHS is a sum of $m$ terms and for each $i$, the factor $\lambda - a_i$
appear in $m-1$ terms (i.e. all terms excluding the $i^{th}$ term ).
At $\lambda = a_i$, they won't contribute. As a result
$$P'(a_i) = \prod_{j=1,\ne i}^m (a_i - a_j)$$
This allow us to rewrite the sum at hand as
$$\mathcal{S}\stackrel{def}{=}\sum_{i=1}^m \frac{1}{\prod_{j=1,\ne i}^m(a_j - a_i)}
= (-1)^{m-1}
\sum_{i=1}^m \frac{1}{\prod_{j=1,\ne i}^m(a_i - a_j)}
= (-1)^{m-1}\sum_{i=1}^m\frac{1}{P'(a_i)}$$
Since the roots of $P(\lambda)$ are distinct, the partial fraction decomposition for $\displaystyle\;\frac{1}{P(\lambda)}$ equals to
$$\frac{1}{P(\lambda)} = \sum_{i=1}^m \frac{1}{P'(a_i)(\lambda - a_i)}\tag{*1}$$
As a consequence,
$$\mathcal{S}
= (-1)^{m-1}\lim_{\lambda\to\infty} \sum_{i=1}^m\frac{\lambda}{P'(a_i)(\lambda-a_i)}
= (-1)^{m-1}\lim_{\lambda\to\infty} \frac{\lambda}{P(\lambda)} = 0
$$
In case one need to justify $(*1)$, multiply RHS$(*1)$ by $P(\lambda)$, one obtain
$$Q(\lambda) \stackrel{def}{=}{\rm RHS}(*1) P(\lambda) = 
\sum_{i=1}^m \frac{1}{P'(a_i)(\lambda - a_i)}\prod_{j=1}^m(\lambda - a_j)
=\sum_{i=1}^m \frac{1}{P'(a_i)} \prod_{j=1,\ne i}^m (\lambda - a_j)
$$
This is a sum of $m$ polynomials in $\lambda$ with degree $m-1$. This means $Q(\lambda)$ is also a polynomial in $\lambda$ with $\deg Q \le m-1$. Once again,
for each $i$, the factor $\lambda - a_i$ appear in all but the $i^{th}$ polynomials. At $\lambda = a_i$, only the $i^{th}$ polynomial contributes and
$$Q(a_i) = \frac{1}{P'(a_i)} \prod_{j=1,\ne i}^m (a_i - a_j) = \frac{P'(a_i)}{P'(a_i)} = 1$$
Since $Q(\lambda) = 1$ at $m > \deg Q$ values of $\lambda$,
$Q(\lambda)$ equals to $1$ identically. This establishes $(*1)$.
A: Given $n$ distinct abscissas $a_i$ and arbitrary ordinates $y_i$, the Lagrange interpolating polynomial is
$$P(x)=\sum_{i=1}^n y_i\dfrac{\prod\limits_{j\ne i} (x-a_j)}{\prod\limits_{j\ne i} (a_i-a_j)}$$
We can choose the $y_i$, and build the corresponding interpolating polynomial $P$. For some $x_0$ such that $x_0\ne a_i$ for all $i$, let
$$\dfrac 1M=\prod\limits_{i=1}^n (x_0-a_i)$$
and
$$y_i=M(x_0-a_i)=\dfrac{1}{\prod\limits_{j\ne i} (x_0-a_j)}$$
We have, on one hand,
$$P(x_0)=\sum_{i=1}^n y_i\dfrac{\prod\limits_{j\ne i} (x_0-a_j)}{\prod\limits_{j\ne i} (a_i-a_j)}=\sum_{i=1}^n \dfrac{1}{\prod\limits_{j\ne i} (a_i-a_j)}$$
On the other hand, since $P$ is an interpolating polynomial, for all $i$,
$$P(a_i)=y_i=M(x_0-a_i)$$
But since the interpolating polynomial is unique among polynomials of degree at most $n-1$ (and $n-1\ge 1$ here, as we assume $n\ge2$), that means that for all $x$,
$$P(x)=M(x_0-x)$$
Then
$$P(x_0)=\sum_{i=1}^n \dfrac{1}{\prod\limits_{j\ne i} (a_i-a_j)}=0$$

Note: the interpolation polynomial is unique because if $P$ and $Q$ are two interpolating polynomials of degree at most $n-1$ for the same abscissas and ordinates, the polynomial $P-Q$, which is also of degree at most $n-1$, has $n$ roots (the $a_i$), hence it's the null polynomial.
