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

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.

Answer 1

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 $\displaystyle P(0)=\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i})$

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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 $\displaystyle P(O)=\sum_{j=1}^k\frac{1}{\alpha_j}$

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

Answer 2

Compare my answer to this question.

The equality is in fact true for all distinct non-zero complex numbers $\alpha_1,\ldots,\alpha_k$, and can be proved as follows. Since I will be using complex numbers I don't want to use $i$ as a variable, so I am slightly changing your notation. Write $$f(z)=\frac{1}{z^2}\prod_{j=1}^k \frac{\alpha_j}{\alpha_j-z}$$ and consider $$\lim_{R\to\infty}\int_{|z|=R} f(z)\,dz\ .$$ First, by estimating $|f(z)|$ on the contour, the limit will be zero. Secondly, we evaluate the integral by residues. The residue at $z=\alpha_m$ is $$\frac{1}{\alpha_m^2}(-\alpha_m)\prod_{j\ne m}\frac{\alpha_j}{\alpha_j-\alpha_m}=-\frac{1}{\alpha_m}\prod_{j\ne m}\frac{\alpha_j}{\alpha_j-\alpha_m}\ .$$ The integrand has a double pole at $z=0$ and we evaluate the residue there using the generalised derivative of a product formula $$\frac{d}{dz}(u_1u_2u_3\cdots)=(u_1u_2u_3\cdots)\Bigl(\frac{u_1'}{u_1}+\frac{u_2'}{u_2}+\frac{u_3'}{u_3}+\cdots\Bigr)\ .$$ The residue is $$\eqalign{ \lim_{z\to0}\frac{d}{dz}\Bigl(\prod_{j=1}^k \frac{\alpha_j}{\alpha_j-z}\Bigr) &=\lim_{z\to0}\Bigl(\prod_{j=1}^k \frac{\alpha_j}{\alpha_j-z}\Bigr) \sum_{j=1}^k\frac{1}{\alpha_j-z} =\sum_{j=1}^k\frac{1}{\alpha_j}\ . \cr}$$ So as long as $R$ is bigger than the maximum of all $|\alpha_j|$, the integral has the constant value $$2\pi i\Bigl(\sum_{j=1}^k\frac{1}{\alpha_j}-\sum_{m=1}^k\frac{1}{\alpha_m}\prod_{j\ne m}\frac{\alpha_j}{\alpha_j-\alpha_m}\Bigr)\ ;$$ this must be zero, and the result follows.