TITLE: How to prove $\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i})=\sum_{i=1}^k\frac{1}{\alpha_i}$?
QUESTION [8 upvotes]: How to prove $\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i})=\sum_{i=1}^k\frac{1}{\alpha_i}$? Where $\alpha_1, \alpha_2,\ldots, \alpha_k$  are $k$ distinct positive numbers.

REPLY [8 votes]: Let me do this with real methods only.
Let $\displaystyle P(X)=\sum_{i=1}^k \frac{1}{\alpha_i}\prod_{j\neq i}^k \frac{\alpha_j-X}{\alpha_j-\alpha_i}$ be the Lagrange interpolating polynomial for $x\to\frac{1}{x}$ at points $\alpha_i$.
Then $\displaystyle P(0)=\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i})$

Let us rewrite $P(0)$ in another way.
Let $Q(X)=XP(X)-1$.
$Q$ has degree $k$ and has $k$ distinct roots: $\alpha_1,\ldots,\alpha_k$.
Denoting $Q$'s leading coefficient as $\lambda$, we have $Q(X)=\lambda\prod_{j=1}^k (X-\alpha_j)$
The trick: consider $\displaystyle \frac{Q'(X)}{Q(X)}=\sum_{j=1}^k\frac{1}{X-\alpha_j}$
Hence $\displaystyle \frac{Q'(0)}{Q(0)}=-\sum_{j=1}^k\frac{1}{\alpha_j}$
But, by the very definition of $Q$, $Q'(0)=P(0)$ and $Q(0)=-1$.
Hence $\displaystyle P(O)=\sum_{j=1}^k\frac{1}{\alpha_j}$

Therefore, $$\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i})=\sum_{i=1}^k\frac{1}{\alpha_i}$$